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the exponent must be n + 3. an = en + 3 13.4 Write an explicit formula for the nth term of the sequence. {9, − 81, 729, − 6,561, 59,049, …} 13.5 Write an explicit formula for the nth term of the sequence. ⎧ ⎨− 3 4 ⎩, − 9 8, − 27 12, − 81 16, − 243 20 ⎫,... ⎬ ⎭ 1454 Chapter 13 Sequences, Probability, and Counting Theory 13.6 Write an explicit formula for the nth term of the sequence. ⎧ ⎨ 1 e2, ⎩ 1 e, 1, e, e2,... ⎫ ⎬ ⎭ Writing the Terms of a Sequence Defined by a Recursive Formula Sequences occur naturally in the growth patterns of nautilus shells, pinecones, tree branches, and many other natural structures. We may see the sequence in the leaf or branch arrangement, the number of petals of a flower, or the pattern of the chambers in a nautilus shell. Their growth follows the Fibonacci sequence, a famous sequence in which each term can be found by adding the preceding two terms. The numbers in the sequence are 1, 1, 2, 3, 5, 8, 13, 21, 34,…. Other examples from the natural world that exhibit the Fibonacci sequence are the Calla Lily, which has just one petal, the Black-Eyed Susan with 13 petals, and different varieties of daisies that may have 21 or 34 petals. Each term of the Fibonacci sequence depends on the terms that come before it. The Fibonacci sequence cannot easily be written using an explicit formula. Instead, we describe the sequence using a recursive formula, a formula that defines the terms of a sequence using previous terms. A recursive formula always has two parts: the value of an initial term (or terms), and an equation defining an in terms of preceding terms. For example, suppose we know the following: We can find the subsequent terms of the sequence using the first term. a1 = 3 an = 2an − 1 − 1, for n ≥ 2 a1 = 3 a2 = 2a1 − 1 = 2(3) − 1 = 5 a3 = 2a2 − 1 = 2(5) − 1 = 9 a4 =
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2a3 − 1 = 2(9) − 1 = 17 So the first four terms of the sequence are {3, 5, 9, 17}. The recursive formula for the Fibonacci sequence states the first two terms and defines each successive term as the sum of the preceding two terms. a1 = 1 a2 = 1 an = an − 1 + an − 2, for n ≥ 3 To find the tenth term of the sequence, for example, we would need to add the eighth and ninth terms. We were told previously that the eighth and ninth terms are 21 and 34, so a10 = a9 + a8 = 34 + 21 = 55 Recursive Formula A recursive formula is a formula that defines each term of a sequence using preceding term(s). Recursive formulas must always state the initial term, or terms, of the sequence. Must the first two terms always be given in a recursive formula? No. The Fibonacci sequence defines each term using the two preceding terms, but many recursive formulas define each term using only one preceding term. These sequences need only the first term to be defined. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1455 Given a recursive formula with only the first term provided, write the first n terms of a sequence. 1. Identify the initial term, a1, which is given as part of the formula. This is the first term. 2. To find the second term, a2, substitute the initial term into the formula for an − 1. Solve. 3. To find the third term, a3, substitute the second term into the formula. Solve. 4. Repeat until you have solved for the nth term. Example 13.5 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first five terms of the sequence defined by the recursive formula. a1 = 9 an = 3an − 1 − 20, for n ≥ 2 Solution The first term is given in the formula. For each subsequent term, we replace an − 1 preceding term. with the value of the a1 = 9 a2 = 3a1 − 20 = 3(9) − 20 = 27 − 20 = 7 a3 = 3a2 − 20 = 3(7) − 20 = 21 − 20 = 1 a4 = 3a3 − 20 = 3(1) − 20 = 3
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− 20 = − 17 a5 = 3a4 − 20 = 3( − 17) − 20 = − 51 − 20 = − 71 The first five terms are {9, 7, 1, – 17, – 71}. See Figure 13.6. Figure 13.6 13.7 Write the first five terms of the sequence defined by the recursive formula. a1 = 2 an = 2an − 1 + 1, for n ≥ 2 1456 Chapter 13 Sequences, Probability, and Counting Theory Given a recursive formula with two initial terms, write the first n terms of a sequence. 1. 2. Identify the initial term, a1, which is given as part of the formula. Identify the second term, a2, which is given as part of the formula. 3. To find the third term, substitute the initial term and the second term into the formula. Evaluate. 4. Repeat until you have evaluated the nth term. Example 13.6 Writing the Terms of a Sequence Defined by a Recursive Formula Write the first six terms of the sequence defined by the recursive formula. a1 = 1 a2 = 2 an = 3an − 1 + 4an − 2, for n ≥ 3 Solution The first two terms are given. For each subsequent term, we replace an − 1 preceding terms. and an − 2 with the values of the two a3 = 3a2 + 4a1 = 3(2) + 4(1) = 10 a4 = 3a3 + 4a2 = 3(10) + 4(2) = 38 a5 = 3a4 + 4a3 = 3(38) + 4(10) = 154 a6 = 3a5 + 4a4 = 3(154) + 4(38) = 614 The first six terms are {1,2,10,38,154,614}. See Figure 13.7. Figure 13.7 13.8 Write the first 8 terms of the sequence defined by the recursive formula. a1 = 0 a2 = 1 a3 = 1 an = an − 1 an − 2 + an − 3, for n ≥ 4 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory Using Factorial Notation The formulas for some sequences include products of consecutive positive integers. n factorial, written as n!, product of the
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positive integers from 1 to n. For example = 24 = 120 1457 is the An example of formula containing a factorial is an = (n + 1)!. The sixth term of the sequence can be found by substituting 6 for n. The factorial of any whole number n is n(n − 1)! We can therefore also think of 5! as 5 ⋅ 4!. a6 = (6 + 1 = 5040 Factorial n factorial is a mathematical operation that can be defined using a recursive formula. The factorial of n, denoted n!, is defined for a positive integer n as(n − 1)(n − 2) ⋯ (2)(1), for n ≥ 2 (13.1) The special case 0! is defined as 0! = 1. Can factorials always be found using a calculator? No. Factorials get large very quickly—faster than even exponential functions! When the output gets too large for the calculator, it will not be able to calculate the factorial. Example 13.7 Writing the Terms of a Sequence Using Factorials Write the first five terms of the sequence defined by the explicit formula an = 5n (n + 2)!. Solution Substitute n = 1, n = 2, and so on in the formula a1 = a2 = a3 = a4 = a5 = 5(1) (1 + 2)! 5(2) (2 + 2)! 5(3) (3 + 2)! 5(4) (4 + 2)! 5(5) (5 + 2)! = 5 3! = 10 4! = 15 5! = 20 6! = 25 10 4 · 3 · 2 · 1 15 5 · 4 · 3 · 2 · 1 20 12 = 1 8 = 1 36 = 5 25,008 1458 Chapter 13 Sequences, Probability, and Counting Theory The first five terms are ⎧ ⎨5 6 ⎩, 5 12, 1 8, 1 36, 5 1,008 ⎫ ⎬. ⎭ Analysis Figure 13.8 shows the graph of the sequence. Notice that, since factorials grow very quickly, the presence of the factorial term in the denominator results in the denominator becoming much larger than the numerator as n increases. This means the quotient gets smaller and, as the plot of the terms shows, the terms are
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decreasing and nearing zero. Figure 13.8 13.9 Write the first five terms of the sequence defined by the explicit formula an = (n + 1)! 2n. Access this online resource for additional instruction and practice with sequences. • Finding Terms in a Sequence (http://openstaxcollege.org/l/findingterms) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1459 13.1 EXERCISES Verbal 1. Discuss the meaning of a sequence. If a finite sequence is defined by a formula, what is its domain? What about an infinite sequence? 2. Describe three ways that a sequence can be defined. 3. Is the ordered set of even numbers an infinite sequence? What about the ordered set of odd numbers? Explain why or why not. 4. What happens to the terms an of a sequence when there is a negative factor in the formula that is raised to a power that includes n? What is the term used to describe this phenomenon? What is a factorial, and how is it denoted? Use an 5. example to illustrate how factorial notation can be beneficial. Algebraic For the following exercises, write the first four terms of the sequence. 6. 7. 8. 9. 10. 11. 12. 13. an = 2 n − 2 an = − 16 n + 1 an = − (−5) n − 1 n an = 2 n3 an = 2n + 1 n3 an = 1.25 ⋅ (−4) n − 1 an = − 4 ⋅ (−6) n − 1 an = n2 2n + 1 14. an = (−10) n + 1 15. an = − ⎛ ⎜ 4 ⋅ ( − 5) ⎝ 5 n − 1 ⎞ ⎟ ⎠ For the following exercises, write the first eight terms of the piecewise sequence. 16. an = ⎧ ⎨ ⎩ n ( − 2) n − 1 (3) − 2 if n is even if n is odd 17. 18. 19. 20. an = ⎧ ⎨ ⎩ if n ≤ 5 n2 2n + 1 n2 − 5 if n >5 an = ⎧ (2n + 1)2 if n is divisible by 4 �
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� 2 ⎩ n if n is not divisible by 4 an = an = ⎧ ⎨ ⎩ −0.6 ⋅ 5 n − 1 2.5 ⋅ ( − 2) if n is prime or 1 n − 1 if n is composite ⎧ ⎨ ⎩ 4(n2 − 2) if n ≤ 3 or n > 6 n2 − 2 4 if 3 < n ≤ 6 For the following exercises, write an explicit formula for each sequence. 21. 22. 23. 24. 25. 4, 7, 12, 19, 28, … −4, 2, − 10, 14, − 34, … 1, 1, 4 3, 2, 16 5, … 1 + e2, 1 − e2 0, 1 − e1 1 + e3, 1 − e3 1 + e4, 1 − e4 1 + e5, … 1 16, … For the following exercises, write the first five terms of the sequence. 26. a1 = 9, an = an − 1 + n 27. a1 = 3, an = (−3)an − 1 28. 29. 30. a1 = − 4, an = a1 = − 1, an = an − 1 + 2n an − 1 − 1 n − 1 (−3) an − 1 − 2 a1 = − 30, an = ⎛ ⎝2 + an − 1 n ⎛ ⎞ ⎠ ⎝ 1 2 ⎞ ⎠ For the following exercises, write the first eight terms of the sequence. 1460 31. 32. 33. Chapter 13 Sequences, Probability, and Counting Theory 4 + n 2n if n is even ⎧ ⎨ ⎩ 3 + n if n is odd a1 = 1 24, a2 = 1, an = ⎛ ⎝2an − 2 ⎛ ⎞ ⎠ ⎝3an − 1 ⎞ ⎠ an = a1 = − 1, a2 = 5, an = an − 2 ⎛ ⎝3 − an − 1 ⎞ ⎠ 49. a1 = 2, a2 = 10, an = 2⎛ ⎝an − 1 + 2⎞ an − 2 ⎠ a1 = 2, an = ⎛
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⎝−an − 1 + 1⎞ ⎠ 2 50. an = 1, an = an − 1 + 8 For the following exercises, write a recursive formula for each sequence. 51. an = (n + 1)! (n − 1)! 34. 35. 36. 37. 38. −2.5, − 5, − 10, − 20, − 40, … −8, − 6, − 3, 1, 6, … 2, 4, 12, 48, 240, … 35, 38, 41, 44, 47, … 15, 3, 3 5, 3 25, 3 125, ⋯ For the following exercises, evaluate the factorial. 39. 6! 40. 41. 42. ⎛ ⎝ 12 6 ⎞ ⎠! 12! 6! 100! 99! For the following exercises, write the first four terms of the sequence. 43. an = n! n2 44. an = 3 ⋅ n! 4 ⋅ n! 45. an = n! n2 − n − 1 46. an = 100 ⋅ n n(n − 1)! Graphical For the following exercises, graph the first five terms of the indicated sequence an = (−1) n n + n 47. 48. This content is available for free at https://cnx.org/content/col11758/1.5 For the following exercises, write an explicit formula for the sequence using the first five points shown on the graph. 52. 53. 54. For the following exercises, write a recursive formula for the sequence using the first five points shown on the graph. Chapter 13 Sequences, Probability, and Counting Theory 1461 55. 56. Technology Follow these steps recursively using a graphing calculator: to evaluate a sequence defined • On the home screen, key in the value for the initial term a1 and press [ENTER]. • Enter the recursive formula by keying in all numerical values given in the formula, along with the key strokes [2ND] ANS for the previous term an − 1. Press [ENTER]. • Continue pressing [ENTER] to calculate the values for each successive term. For the following exercises, use the steps above to find the indicated term or terms for the sequence. 57. Find a1 = 87 111 give fractional results. the first, an = 4 3 five an − 1 + 12 37
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of terms the sequence. Use the >Frac feature to 59. Find a1 = 2, the first five terms of the sequence an = 2 [(an − 1) − 1] + 1. 60. Find the a1 = 8, an = ⎛ ten first ⎝an − 1 + 1⎞ ⎠! an − 1!. terms of the sequence Find 61. a1 = 2, an = nan − 1 the tenth term of the sequence Follow these steps to evaluate a finite sequence defined by an explicit formula. Using a TI-84, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • • • • In the line headed “Expr:” type in the explicit formula, using the [X,T, θ, n] button for n In the line headed “Variable:” type in the variable used on the previous step. In the line headed “start:” key in the value of n that begins the sequence. In the line headed “end:” key in the value of n that ends the sequence. • Press [ENTER] 3 times to return to the home screen. You will see the sequence syntax on the screen. Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. Using a TI-83, do the following. • In the home screen, press [2ND] LIST. • Scroll over to OPS and choose “seq(” from the dropdown list. Press [ENTER]. • Enter the items in the order “Expr”, “Variable”, “start”, “end” separated by commas. See the instructions above for the description of each item. • Press [ENTER] to see the list of terms for the finite sequence defined. Use the right arrow key to scroll through the list of terms. For the following exercises, use the steps above to find the indicated terms for the sequence. Round to the nearest thousandth when necessary. first List 62. an = − 28 9 the n + 5 3. five terms of the sequence 63. List six the first terms n3 − 3.5n2 + 4.1n − 1.5
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2.4n. of the sequence 58. Find a1 = 625, 15th the an = 0.8an − 1 + 18. term of the sequence an = 64. 1462 List an = five terms of the sequence Chapter 13 Sequences, Probability, and Counting Theory the first 15n ⋅ (−2) n − 1 47 65. List an = 5.7 n the four first + 0.275(n − 1)! terms of the sequence 66. List the first six terms of the sequence an = n! n. Extensions Consider the sequence defined by an = − 6 − 8n. Is 67. an = − 421 a term in the sequence? Verify the result. 68. What term in the sequence an = n2 + 4n + 4 2(n + 2) has the value 41? Verify the result. a Find recursive sequence 69. 1, 0, − 1, − 1, 0, 1, 1, 0, − 1, − 1, 0, 1, 1,.... (Hint: find a pattern for an based on the first two terms.) formula the for 70. Calculate the first eight terms of the sequences and bn = n3 + 3n2 + 2n, and then make the relationship between these two an = (n + 2)! (n − 1)! a conjecture about sequences. 71. Prove the conjecture made in the preceding exercise. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1463 13.2 | Arithmetic Sequences Learning Objectives In this section, you will: 13.2.1 Find the common difference for an arithmetic sequence. 13.2.2 Write terms of an arithmetic sequence. 13.2.3 Use a recursive formula for an arithmetic sequence. 13.2.4 Use an explicit formula for an arithmetic sequence. Companies often make large purchases, such as computers and vehicles, for business use. The book-value of these supplies decreases each year for tax purposes. This decrease in value is called depreciation. One method of calculating depreciation is straight-line depreciation, in which the value of the asset decreases by the same amount each year. As an example, consider a woman who starts a small contracting business. She purchases a new truck for $25,000. After five years, she estimates that she will be able to sell the
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truck for $8,000. The loss in value of the truck will therefore be $17,000, which is $3,400 per year for five years. The truck will be worth $21,600 after the first year; $18,200 after two years; $14,800 after three years; $11,400 after four years; and $8,000 at the end of five years. In this section, we will consider specific kinds of sequences that will allow us to calculate depreciation, such as the truck’s value. Finding Common Differences The values of the truck in the example are said to form an arithmetic sequence because they change by a constant amount each year. Each term increases or decreases by the same constant value called the common difference of the sequence. For this sequence, the common difference is –3,400. The sequence below is another example of an arithmetic sequence. In this case, the constant difference is 3. You can choose any term of the sequence, and add 3 to find the subsequent term. Arithmetic Sequence An arithmetic sequence is a sequence that has the property that the difference between any two consecutive terms is a constant. This constant is called the common difference. If a1 is the first term of an arithmetic sequence and d is the common difference, the sequence will be: ⎧ ⎨an ⎩ ⎫ ⎬ = ⎭ ⎧ ⎨a1, a1 + d, a1 + 2d, a1 + 3d,... ⎩ ⎫ ⎬ ⎭ Example 13.8 Finding Common Differences Is each sequence arithmetic? If so, find the common difference. a. b. {1, 2, 4, 8, 16,...} { − 3, 1, 5, 9, 13,...} 1464 Chapter 13 Sequences, Probability, and Counting Theory Solution Subtract each term from the subsequent term to determine whether a common difference exists. a. The sequence is not arithmetic because there is no common difference. b. The sequence is arithmetic because there is a common difference. The common difference is 4. Analysis The graph of each of these sequences is shown in Figure 13.9. We can see from the graphs that, although both sequences show growth, a is not linear whereas b is linear. Arithmetic sequences have a constant rate of change so their graphs will always be points on a line. Figure 13.9 If we are told that a sequence
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is arithmetic, do we have to subtract every term from the following term to find the common difference? No. If we know that the sequence is arithmetic, we can choose any one term in the sequence, and subtract it from the subsequent term to find the common difference. 13.10 Is the given sequence arithmetic? If so, find the common difference. {18, 16, 14, 12, 10, … } 13.11 Is the given sequence arithmetic? If so, find the common difference. {1, 3, 6, 10, 15, … } Writing Terms of Arithmetic Sequences Now that we can recognize an arithmetic sequence, we will find the terms if we are given the first term and the common difference. The terms can be found by beginning with the first term and adding the common difference repeatedly. In addition, any term can also be found by plugging in the values of n and d into formula below. an = a1 + (n − 1)d This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1465 Given the first term and the common difference of an arithmetic sequence, find the first several terms. 1. Add the common difference to the first term to find the second term. 2. Add the common difference to the second term to find the third term. 3. Continue until all of the desired terms are identified. 4. Write the terms separated by commas within brackets. Example 13.9 Writing Terms of Arithmetic Sequences Write the first five terms of the arithmetic sequence with a1 = 17 and d = − 3. Solution Adding − 3 is the same as subtracting 3. Beginning with the first term, subtract 3 from each term to find the next term. The first five terms are {17, 14, 11, 8, 5} Analysis As expected, the graph of the sequence consists of points on a line as shown in Figure 13.10. Figure 13.10 13.12 List the first five terms of the arithmetic sequence with a1 = 1 and d = 5. Given any the first term and any other term in an arithmetic sequence, find a given term. 1. Substitute the values given for a1, an, n into the formula an = a1 + (n − 1)d to solve for d. 2. Find a given term by substituting the appropriate values for a1, n, and d into the formula
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an = a1 + (n − 1)d. Example 13.10 Writing Terms of Arithmetic Sequences 1466 Chapter 13 Sequences, Probability, and Counting Theory Given a1 = 8 and a4 = 14, find a5. Solution The sequence can be written in terms of the initial term 8 and the common difference d. {8, 8 + d, 8 + 2d, 8 + 3d} We know the fourth term equals 14; we know the fourth term has the form a1 + 3d = 8 + 3d. We can find the common difference d. an = a1 + (n − 1)d a4 = a1 + 3d a4 = 8 + 3d 14 = 8 + 3d d = 2 Write the fourth term of the sequence in terms of a1 and d. Substitute 14 for a4. Solve for the common diffe ence. Find the fifth term by adding the common difference to the fourth term. a5 = a4 + 2 = 16 Analysis Notice that the common difference is added to the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by adding the common difference to the first term nine times or by using the equation an = a1 + (n − 1)d. 13.13 Given a3 = 7 and a5 = 17, find a2. Using Recursive Formulas for Arithmetic Sequences Some arithmetic sequences are defined in terms of the previous term using a recursive formula. The formula provides an algebraic rule for determining the terms of the sequence. A recursive formula allows us to find any term of an arithmetic sequence using a function of the preceding term. Each term is the sum of the previous term and the common difference. For example, if the common difference is 5, then each term is the previous term plus 5. As with any recursive formula, the first term must be given. an = an − 1 + d n ≥ 2 Recursive Formula for an Arithmetic Sequence The recursive formula for an arithmetic sequence with common difference d is: an = an − 1 + d n ≥ 2 (13.2) Given an arithmetic sequence, write its recursive formula. 1. Subtract any term from the subsequent term to find the common difference. 2. State the initial term and substitute the common difference into the recursive formula for arithmetic sequences. This content is available for free at
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ERROR: type should be string, got " https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1467 Example 13.11 Writing a Recursive Formula for an Arithmetic Sequence Write a recursive formula for the arithmetic sequence. { − 18, − 7, 4, 15, 26, …} Solution The first term is given as −18. The common difference can be found by subtracting the first term from the second term. Substitute the initial term and the common difference into the recursive formula for arithmetic sequences. d = −7 − (−18) = 11 a1 = − 18 an = an − 1 + 11, for n ≥ 2 Analysis We see that the common difference is the slope of the line formed when we graph the terms of the sequence, as shown in Figure 13.11. The growth pattern of the sequence shows the constant difference of 11 units. Figure 13.11 Do we have to subtract the first term from the second term to find the common difference? No. We can subtract any term in the sequence from the subsequent term. It is, however, most common to subtract the first term from the second term because it is often the easiest method of finding the common difference. 13.14 Write a recursive formula for the arithmetic sequence. {25, 37, 49, 61, …} Using Explicit Formulas for Arithmetic Sequences We can think of an arithmetic sequence as a function on the domain of the natural numbers; it is a linear function because it has a constant rate of change. The common difference is the constant rate of change, or the slope of the function. We can construct the linear function if we know the slope and the vertical intercept. To find the y-intercept of the function, we can subtract the common difference from the first term of the sequence. Consider the following sequence. an = a1 + d(n − 1) 1468 Chapter 13 Sequences, Probability, and Counting Theory The common difference is −50, so the sequence represents a linear function with a slope of −50. To find the y intercept, we subtract −50 from 200 : 200 − ( − 50) = 200 + 50 = 250. You can also find the y -intercept by graphing the function and determining where a line that connects the points would intersect the vertical axis. The graph is shown in Figure 13.12. Figure 13.12 Recall the slope-intercept form of a line is y = mx + b. When" |
dealing with sequences, we use an in place of y and n in place of x. If we know the slope and vertical intercept of the function, we can substitute them for m and b in the slopeintercept form of a line. Substituting − 50 for the slope and 250 for the vertical intercept, we get the following equation: an = − 50n + 250 We do not need to find the vertical intercept to write an explicit formula for an arithmetic sequence. Another explicit formula for this sequence is an = 200 − 50(n − 1), which simplifies to an = − 50n + 250. Explicit Formula for an Arithmetic Sequence An explicit formula for the nth term of an arithmetic sequence is given by an = a1 + d(n − 1) (13.3) Given the first several terms for an arithmetic sequence, write an explicit formula. 1. Find the common difference, a2 − a1. 2. Substitute the common difference and the first term into an = a1 + d(n − 1). Example 13.12 Writing the nth Term Explicit Formula for an Arithmetic Sequence Write an explicit formula for the arithmetic sequence. {2, 12, 22, 32, 42, …} Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1469 The common difference can be found by subtracting the first term from the second term. d = a2 − a1 = 12 − 2 = 10 The common difference is 10. Substitute the common difference and the first term of the sequence into the formula and simplify. an = 2 + 10(n − 1) an = 10n − 8 Analysis The graph of this sequence, represented in Figure 13.13, shows a slope of 10 and a vertical intercept of −8. Figure 13.13 13.15 Write an explicit formula for the following arithmetic sequence. {50, 47, 44, 41, … } Finding the Number of Terms in a Finite Arithmetic Sequence Explicit formulas can be used to determine the number of terms in a finite arithmetic sequence. We need to find the common difference, and then determine how many times the common difference must be added to the first term to obtain the final term of the sequence. Given the first three terms and the last term of a finite arithmetic sequence, find the total number of terms. 1. Find the common difference d. 2. Substitute
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the common difference and the first term into an = a1 + d(n – 1). 3. Substitute the last term for an and solve for n. Example 13.13 Finding the Number of Terms in a Finite Arithmetic Sequence Find the number of terms in the finite arithmetic sequence. {8, 1, –6,..., –41} 1470 Chapter 13 Sequences, Probability, and Counting Theory Solution The common difference can be found by subtracting the first term from the second term. 1 − 8 = − 7 The common difference is −7. Substitute the common difference and the initial term of the sequence into the nth term formula and simplify. an = a1 + d(n − 1) an = 8 + − 7(n − 1) an = 15 − 7n −41 = 15 − 7n 8 = n Substitute −41 for an and solve for n There are eight terms in the sequence. 13.16 Find the number of terms in the finite arithmetic sequence. {6, 11, 16,..., 56} Solving Application Problems with Arithmetic Sequences In many application problems, it often makes sense to use an initial term of a0 the explicit formula slightly to account for the difference in initial terms. We use the following formula: instead of a1. In these problems, we alter an = a0 + dn Example 13.14 Solving Application Problems with Arithmetic Sequences A five-year old child receives an allowance of $1 each week. His parents promise him an annual increase of $2 per week. a. Write a formula for the child’s weekly allowance in a given year. b. What will the child’s allowance be when he is 16 years old? Solution a. The situation can be modeled by an arithmetic sequence with an initial term of 1 and a common difference of 2. Let A be the amount of the allowance and n be the number of years after age 5. Using the altered explicit formula for an arithmetic sequence we get: An = 1 + 2n b. We can find the number of years since age 5 by subtracting. 16 − 5 = 11 We are looking for the child’s allowance after 11 years. Substitute 11 into the formula to find the child’s allowance at age 16. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 14
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71 The child’s allowance at age 16 will be $23 per week. A11 = 1 + 2(11) = 23 13.17 A woman decides to go for a 10-minute run every day this week and plans to increase the time of her daily run by 4 minutes each week. Write a formula for the time of her run after n weeks. How long will her daily run be 8 weeks from today? Access this online resource for additional instruction and practice with arithmetic sequences. • Arithmetic Sequences (http://openstaxcollege.org/l/arithmeticseq) 1472 Chapter 13 Sequences, Probability, and Counting Theory 13.2 EXERCISES Verbal 72. What is an arithmetic sequence? How is the common difference of an arithmetic 73. sequence found? How do we determine whether a sequence is 74. arithmetic? 75. What are the main differences between using a recursive formula and using an explicit formula to describe an arithmetic sequence? Describe 76. and how linear sequences are similar. How are they different? functions arithmetic Algebraic For the following exercises, find the common difference for the arithmetic sequence provided. 77. 78. {5, 11, 17, 23, 29,...} ⎧ ⎨0, 1 2 ⎩, 1, 3 2 ⎫, 2,... ⎬ ⎭ the following exercises, determine whether For sequence is arithmetic. If so find the common difference. the 79. 80. {11.4, 9.3, 7.2, 5.1, 3,...} {4, 16, 64, 256, 1024,...} For the following exercises, write the first five terms of the arithmetic sequence given the first term and common difference. 81. a1 = −25, d = −9 82. a1 = 0, d = 2 3 For the following exercises, write the first five terms of the arithmetic series given two terms. 83. a1 = 17, a7 = − 31 84. a13 = − 60, a33 = − 160 For the following exercises, find the specified term for the arithmetic sequence given the first term and common difference. First term is 3, common difference is 4, find the 5th 85. term. 86. This content is available for free at https://cnx.org/content/col11758/1.5 First term is 4, common difference is 5, find the 4th term. First term
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is 5, common difference is 6, find the 8th 87. term. First term is 6, common difference is 7, find the 6th 88. term. First term is 7, common difference is 8, find the 7th 89. term. For the following exercises, find the first term given two terms from an arithmetic sequence. Find the first term or a1 90. a6 = 12 and a14 = 28. Find the first term or a1 91. a7 = 21 and a15 = 42. Find the first term or a1 92. a8 = 40 and a23 = 115. Find the first term or a1 93. a9 = 54 and a17 = 102. Find the first term or a1 94. a11 = 11 and a21 = 16. of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if of an arithmetic sequence if For the following exercises, find the specified term given two terms from an arithmetic sequence. 95. a1 = 33 and a7 = − 15. Find a4. 96. a3 = − 17.1 and a10 = − 15.7. Find a21. For the following exercises, use the recursive formula to write the first five terms of the arithmetic sequence. 97. 98. a1 = 39; an = an − 1 − 3 a1 = − 19; an = an − 1 − 1.4 For the following exercises, write a recursive formula for each arithmetic sequence. 99. an = {40, 60, 80,...} 100. 101. an = ⎧ ⎨17, 26, 35,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨ − 1, 2, 5,...⎫ ⎬ ⎭ ⎩ Chapter 13 Sequences, Probability, and Counting Theory 1473 102. an = {12, 17, 22,...} 123. an = ⎧ ⎨−5, − 10 3 ⎩, − 5 3, … ⎫ ⎬ ⎭ 103. an = ⎧ ⎨ − 15, − 7, 1,...⎫ ⎬ ⎭ ⎩ 104. an = {8.9, 10.3, 11.7,...} For the following exercises, find the number of terms in the given finite arithmetic sequence. 105
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. 106. an = ⎧ ⎨ − 0.52, − 1.02, − 1.52,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 5 ⎩, 9 20, 7 10 ⎫,... ⎬ ⎭ 107. an = ⎧ ⎨−,... ⎬ ⎭ 108. an = ⎧ ⎨1 6 ⎩, − 11 12 ⎫, − 2,... ⎬ ⎭ 124. 125. 126. an = ⎧ ⎨3, − 4, − 11,..., − 60⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1.2, 1.4, 1.6,..., 3.8⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 2 ⎩, 2, 7 2,..., 8 ⎫ ⎬ ⎭ Graphical For the following exercises, determine whether the graph shown represents an arithmetic sequence. For the following exercises, write a recursive formula for the given arithmetic sequence, and then find the specified term. 127. 109. an = {7, 4, 1,...}; Find the 17th term. 110. an = {4, 11, 18,...}; Find the 14th term. 111. an = ⎧ ⎨2, 6, 10,...⎫ ⎭ ⎩ ⎬; Find the 12th term. For the following exercises, use the explicit formula to write the first five terms of the arithmetic sequence. 112. an = 24 − 4n 113. an = 1 2 n − 1 2 For the following exercises, write an explicit formula for each arithmetic sequence. 114. 115. 116. an = ⎧ ⎨3, 5, 7,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨32, 24, 16,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨ − 5, 95, 195,...⎫ ⎬ ⎭ ⎩ 117. an = {−17, −217, −417,...} 118. 119. 120. 121. an = ⎧
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⎨1.8, 3.6, 5.4,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨−18.1, −16.2, −14.3,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨15.8, 18.5, 21.2,...⎫ ⎬ ⎭ ⎩ an = ⎧ ⎨1 3 ⎩, − 4 3, −3,... ⎫ ⎬ ⎭ 122. an = ⎧ ⎨0, 1 3 ⎩ ⎫,... ⎬ ⎭, 2 3 128. 1474 Chapter 13 Sequences, Probability, and Counting Theory ◦ Set TblStart = 1 ◦ Set ΔTbl = 1 ◦ Set Indpnt: Auto and Depend: Auto • Press [2ND] then [GRAPH] to go to the TABLE What are the first seven terms shown in the column 132. with the heading u(n)? Use the scroll-down arrow to scroll to n = 50. What 133. value is given for u(n)? Press [WINDOW]. 134. nMin = 1, nMax = 5, xMin = 0, xMax = 6, yMin = − 1, and yMax = 14. Then press sequence as it appears on the graphing calculator. [GRAPH]. Graph Set the For the following exercises, follow the steps given above to work with the arithmetic sequence an = 1 n + 5 using a 2 graphing calculator. What are the first seven terms shown in the column 135. with the heading u(n) in the TABLE feature? 136. Graph the sequence as it appears on the graphing calculator. Be sure to adjust the WINDOW settings as needed. Extensions Give two examples of arithmetic sequences whose 4th 137. terms are 9. Give two examples of arithmetic sequences whose 138. 10th terms are 206. For the following exercises, use the information provided to graph the first 5 terms of the arithmetic sequence. 129. a1 = 0, d = 4 130. a1 = 9; an = an − 1 − 10 131. an = − 12 + 5n Technology For the following exercises, follow the steps to work with the arithmetic sequence an = 3n − 2 using a graphing calculator: Find
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the 5th term of 139. {9b, 5b, b, … }. the arithmetic sequence • Press [MODE] ◦ Select SEQ in the fourth line ◦ Select DOT in the fifth line ◦ Press [ENTER] • Press [Y=] ◦ ◦ ◦ nMin is the first counting number for the sequence. Set nMin = 1 u(n) is the pattern for the sequence. Set u(n) = 3n − 2 u(nMin) is the sequence. Set u(nMin) = 1 first number in the • Press [2ND] then [WINDOW] to go to TBLSET This content is available for free at https://cnx.org/content/col11758/1.5 Find the 11th term of 140. {3a − 2b, a + 2b, − a + 6b … }. the arithmetic sequence At 141. {5.4, 14.5, 23.6,...} exceed 151? which term does the sequence 142. At which term does the sequence ⎧ ⎨17 3 ⎩, 31 6, 14 3 ⎫,... ⎬ ⎭ begin to have negative values? For which terms does the finite arithmetic sequence 143. ⎧ ⎨5 2 ⎩, 19 8, 9 4,..., 1 8 ⎫ ⎬ have integer values? ⎭ Write an arithmetic sequence using a recursive 144. formula. Show the first 4 terms, and then find the 31st term. Chapter 13 Sequences, Probability, and Counting Theory 1475 145. Write an arithmetic sequence using an explicit formula. Show the first 4 terms, and then find the 28th term. 1476 Chapter 13 Sequences, Probability, and Counting Theory 13.3 | Geometric Sequences Learning Objectives In this section, you will: 13.3.1 Find the common ratio for a geometric sequence. 13.3.2 List the terms of a geometric sequence. 13.3.3 Use a recursive formula for a geometric sequence. 13.3.4 Use an explicit formula for a geometric sequence. Many jobs offer an annual cost-of-living increase to keep salaries consistent with inflation. Suppose, for example, a recent college graduate finds a position as a sales manager earning an annual salary of $26,000. He is promised a 2% cost of living increase each year
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. His annual salary in any given year can be found by multiplying his salary from the previous year by 102%. His salary will be $26,520 after one year; $27,050.40 after two years; $27,591.41 after three years; and so on. When a salary increases by a constant rate each year, the salary grows by a constant factor. In this section, we will review sequences that grow in this way. Finding Common Ratios The yearly salary values described form a geometric sequence because they change by a constant factor each year. Each term of a geometric sequence increases or decreases by a constant factor called the common ratio. The sequence below is an example of a geometric sequence because each term increases by a constant factor of 6. Multiplying any term of the sequence by the common ratio 6 generates the subsequent term. Definition of a Geometric Sequence A geometric sequence is one in which any term divided by the previous term is a constant. This constant is called the common ratio of the sequence. The common ratio can be found by dividing any term in the sequence by the previous term. If a1 is the initial term of a geometric sequence and r is the common ratio, the sequence will be ⎧ ⎨a1, ⎩ a1 r, a1 r 2, a1 r 3,... ⎫ ⎬. ⎭ Given a set of numbers, determine if they represent a geometric sequence. 1. Divide each term by the previous term. 2. Compare the quotients. If they are the same, a common ratio exists and the sequence is geometric. Example 13.15 Finding Common Ratios Is the sequence geometric? If so, find the common ratio. a. b. 1, 2, 4, 8, 16,... 48, 12, 4, 2,... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1477 Solution Divide each term by the previous term to determine whether a common ratio exists. a. b. = 2 2 1 The sequence is geometric because there is a common ratio. The common ratio is 2. 16 12 48 The sequence is not geometric because there is not a common ratio. = 1 3 = 1 2 4 12 2 4 Analysis The graph of each sequence is shown in Figure 13.14. It seems from the graphs that both (a) and (b) appear have the form
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of the graph of an exponential function in this viewing window. However, we know that (a) is geometric and so this interpretation holds, but (b) is not. Figure 13.14 If you are told that a sequence is geometric, do you have to divide every term by the previous term to find the common ratio? No. If you know that the sequence is geometric, you can choose any one term in the sequence and divide it by the previous term to find the common ratio. 13.18 Is the sequence geometric? If so, find the common ratio. 5, 10, 15, 20,... 13.19 Is the sequence geometric? If so, find the common ratio. 100, 20, 4, 4 5,... Writing Terms of Geometric Sequences Now that we can identify a geometric sequence, we will learn how to find the terms of a geometric sequence if we are given the first term and the common ratio. The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. For instance, if the first term of a geometric sequence is a1 = − 2 and the common ratio is r = 4, we can find subsequent terms by multiplying −2 ⋅ 4 to get −8 then multiplying the result −8 ⋅ 4 to get −32 and so on. 1478 Chapter 13 Sequences, Probability, and Counting Theory a1 = − 2 a2 = ( − 2 ⋅ 4) = − 8 a3 = ( − 8 ⋅ 4) = − 32 a4 = ( − 32 ⋅ 4) − 128 The first four terms are {–2, –8, –32, –128}. Given the first term and the common factor, find the first four terms of a geometric sequence. 1. Multiply the initial term, a1, by the common ratio to find the next term, a2. 2. Repeat the process, using an = a2 to find a3 and then a3 to find a4, until all four terms have been identified. 3. Write the terms separated by commons within brackets. Example 13.16 Writing the Terms of a Geometric Sequence List the first four terms of the geometric sequence with a1 = 5 and r = –2. Solution Multiply a1 by −2 to find a2. Repeat the process, using a2 to find a3, and so on. a1 = 5 a2 = − 2a
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1 = − 10 a3 = − 2a2 = 20 a4 = − 2a3 = − 40 The first four terms are {5, –10, 20, –40}. 13.20 List the first five terms of the geometric sequence with a1 = 18 and r = 1 3. Using Recursive Formulas for Geometric Sequences A recursive formula allows us to find any term of a geometric sequence by using the previous term. Each term is the product of the common ratio and the previous term. For example, suppose the common ratio is 9. Then each term is nine times the previous term. As with any recursive formula, the initial term must be given. Recursive Formula for a Geometric Sequence The recursive formula for a geometric sequence with common ratio r and first term a1 is an = ran − 1, n ≥ 2 (13.4) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1479 Given the first several terms of a geometric sequence, write its recursive formula. 1. State the initial term. 2. Find the common ratio by dividing any term by the preceding term. 3. Substitute the common ratio into the recursive formula for a geometric sequence. Example 13.17 Using Recursive Formulas for Geometric Sequences Write a recursive formula for the following geometric sequence. {6, 9, 13.5, 20.25,...} Solution The first term is given as 6. The common ratio can be found by dividing the second term by the first term. Substitute the common ratio into the recursive formula for geometric sequences and define a1. r = 9 6 = 1.5 an = ran − 1 an = 1.5an − 1 for n ≥ 2 a1 = 6 Analysis The sequence of data points follows an exponential pattern. The common ratio is also the base of an exponential function as shown in Figure 13.15 Figure 13.15 Do we have to divide the second term by the first term to find the common ratio? No. We can divide any term in the sequence by the previous term. It is, however, most common to divide the second term by the first term because it is often the easiest method of finding the common ratio. 13.21 Write a recursive formula for the following geometric sequence. ⎧ ⎨2, 4 3 ⎩, 8 9, 16 27,... ⎫ �
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� ⎭ Using Explicit Formulas for Geometric Sequences Because a geometric sequence is an exponential function whose domain is the set of positive integers, and the common ratio is the base of the function, we can write explicit formulas that allow us to find particular terms. 1480 Chapter 13 Sequences, Probability, and Counting Theory an = a1 r n − 1 Let’s take a look at the sequence {18, 36, 72, 144, 288,...}. This is a geometric sequence with a common ratio of 2 and an exponential function with a base of 2. An explicit formula for this sequence is The graph of the sequence is shown in Figure 13.16. an = 18 · 2 n − 1 Figure 13.16 Explicit Formula for a Geometric Sequence The nth term of a geometric sequence is given by the explicit formula: an = a1 r n − 1 (13.5) Example 13.18 Writing Terms of Geometric Sequences Using the Explicit Formula Given a geometric sequence with a1 = 3 and a4 = 24, find a2. Solution The sequence can be written in terms of the initial term and the common ratio r. Find the common ratio using the given fourth term. 3, 3r, 3r 2, 3r 3,... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1481 an = a1 r n − 1 a4 = 3r 3 24 = 3r 3 8 = r 3 r = 2 Write the fourth term of sequence in terms of α1 and r Substitute 24 for a4 Divide Solve for the common ratio Find the second term by multiplying the first term by the common ratio. a2 = 2a1 = 2(3) = 6 Analysis The common ratio is multiplied by the first term once to find the second term, twice to find the third term, three times to find the fourth term, and so on. The tenth term could be found by multiplying the first term by the common ratio nine times or by multiplying by the common ratio raised to the ninth power. 13.22 Given a geometric sequence with a2 = 4 and a3 = 32, find a6. Example 13.19 Writing an Explicit Formula for the nth Term of a Geometric Sequence Write an explicit formula for the nth term of the following geometric sequence. {2, 10, 50, 250,...
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} Solution The first term is 2. The common ratio can be found by dividing the second term by the first term. The common ratio is 5. Substitute the common ratio and the first term of the sequence into the formula. 10 2 = 5 The graph of this sequence in Figure 13.17 shows an exponential pattern. an = a1 r (n − 1) n − 1 an = 2 ⋅ 5 1482 Chapter 13 Sequences, Probability, and Counting Theory Figure 13.17 13.23 Write an explicit formula for the following geometric sequence. {–1, 3, –9, 27,...} Solving Application Problems with Geometric Sequences In real-world scenarios involving arithmetic sequences, we may need to use an initial term of a0 problems, we can alter the explicit formula slightly by using the following formula: instead of a1. In these an = a0 r n Example 13.20 Solving Application Problems with Geometric Sequences In 2013, the number of students in a small school is 284. It is estimated that the student population will increase by 4% each year. a. Write a formula for the student population. b. Estimate the student population in 2020. Solution a. The situation can be modeled by a geometric sequence with an initial term of 284. The student population will be 104% of the prior year, so the common ratio is 1.04. Let P be the student population and n be the number of years after 2013. Using the explicit formula for a geometric sequence we get b. We can find the number of years since 2013 by subtracting. Pn = 284 ⋅ 1.04 n 2020 − 2013 = 7 We are looking for the population after 7 years. We can substitute 7 for n to estimate the population in 2020. P7 = 284 ⋅ 1.047 ≈ 374 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1483 The student population will be about 374 in 2020. A business starts a new website. Initially the number of hits is 293 due to the curiosity factor. The 13.24 business estimates the number of hits will increase by 2.6% per week. a. Write a formula for the number of hits. b. Estimate the number of hits in 5 weeks. Access these online resources for additional instruction and practice with geometric sequences. • Geometric Sequences (http
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://openstaxcollege.org/l/geometricseq) • Determine the Type of Sequence (http://openstaxcollege.org/l/sequencetype) • Find the Formula for a Sequence (http://openstaxcollege.org/l/sequenceformula) 1484 Chapter 13 Sequences, Probability, and Counting Theory 13.3 EXERCISES Verbal 146. What is a geometric sequence? How is the common ratio of a geometric sequence 147. found? What is the procedure for determining whether a 148. sequence is geometric? What 149. sequence and a geometric sequence? is the difference between an arithmetic For the following exercises, find the specified term for the geometric sequence, given the first term and common ratio. The first term is 2, and the common ratio is 3. Find 163. the 5th term. 164. The first term is 16 and the common ratio is − 1 3. Find the 4th term. For the following exercises, find the specified term for the geometric sequence, given the first four terms. Describe how exponential functions and geometric 150. sequences are similar. How are they different? 165. an = {−1, 2, − 4, 8,...}. Find a12. Algebraic For the following exercises, find the common ratio for the geometric sequence. 151. 152. 153. 1, 3, 9, 27, 81,... −0.125, 0.25, − 0.5, 1, − 2,... −2 32, − 1 128,... 166. an = ⎧ ⎨−2 27 ⎫,... ⎭ ⎬. Find a7. For the following exercises, write the first five terms of the geometric sequence. 167. a1 = − 486, an = − 1 3 an − 1 168. a1 = 7, an = 0.2an − 1 the following exercises, determine whether For sequence is geometric. If so, find the common ratio. the For the following exercises, write a recursive formula for each geometric sequence. 154. 155. 156. 157. 158. −6, − 12, − 24, − 48, − 96,... 5, 5.2, 5.4, 5.6, 5.8,... −1 16,... 6, 8, 11, 15, 20,... 0.8, 4, 20, 100, 500,... For the following exercises, write
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the first five terms of the geometric sequence, given the first term and common ratio. 159. a1 = 8, r = 0.3 160. a1 = 5, r = 1 5 169. an = {−1, 5, − 25, 125,...} 170. an = {−32, − 16, − 8, − 4,...} 171. an = {14, 56, 224, 896,...} 172. an = {10, − 3, 0.9, − 0.27,...} 173. an = {0.61, 1.83, 5.49, 16.47,...} 174. an = ⎧ ⎨3 5 ⎩, 1 10, 1 60, 1 360 ⎫,... ⎬ ⎭ 175. an = ⎧ ⎨−2, 4 3 ⎩, − 8 9, 16 27 ⎫,... ⎬ ⎭ 176. an = ⎧ ⎨ 1 512 ⎩, − 1 128, 1 32, − 1 8 ⎫,... ⎬ ⎭ For the following exercises, write the first five terms of the geometric sequence, given any two terms. For the following exercises, write the first five terms of the geometric sequence. 161. a7 = 64, a10 = 512 162. a6 = 25, a8 = 6.25 177. an = − 4 ⋅ 5 n − 1 178. an = 12 ⋅ n − 1 ⎛ ⎝− 1 2 ⎞ ⎠ This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1485 For the following exercises, write an explicit formula for each geometric sequence. 179. an = {−2, − 4, − 8, − 16,...} 180. an = {1, 3, 9, 27,...} 181. an = {−4, − 12, − 36, − 108,...} 182. an = {0.8, − 4, 20, − 100,...} 183. 184. an = ⎧ ⎨ − 1.25, − 5, − 20, − 80,...⎫ ⎬ ⎭ ⎩ an = ⎧ �
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�−1, − 4 5 ⎩, − 16 25, − 64 125 ⎫,... ⎬ ⎭ 185. an = 186. an = ⎧ ⎨2, 1 3 ⎩, 1 18, 1 108 ⎫,... ⎬ ⎭ 192. ⎧ ⎨3, − 1, 1 3 ⎩, − 1 9 ⎫,... ⎬ ⎭ For the following exercises, find the specified term for the geometric sequence given. 187. Let a1 = 4, an = − 3an − 1. Find a8. 188. Let an = − n − 1 ⎛ ⎝− 1 3 ⎞ ⎠. Find a12. For the following exercises, find the number of terms in the given finite geometric sequence. 189. an = {−1, 3, − 9,..., 2187} ⎧ ⎨2, 1, 1 2 ⎩,..., 1 1024 ⎫ ⎬ ⎭ 190. an = Graphical For the following exercises, determine whether the graph shown represents a geometric sequence. 191. For the following exercises, use the information provided to graph the first five terms of the geometric sequence. 193. a1 = 1, r = 1 2 194. a1 = 3, an = 2an − 1 195. an = 27 ⋅ 0.3 n − 1 Extensions 196. 1486 Chapter 13 Sequences, Probability, and Counting Theory Use recursive formulas to give two examples of geometric sequences whose 3rd terms are 200. Use explicit 197. geometric sequences whose 7th terms are 1024. formulas to give two examples of Find the 5th term of 198. {b, 4b, 16b,...}. the geometric sequence Find the 7th term of 199. {64a( − b), 32a( − 3b), 16a( − 9b),...}. the geometric sequence At 200. {10, 12, 14.4, 17.28,...} exceed 100? which does term the sequence 201. ⎧ ⎨ 1 2187 ⎩ At, 1 729 which, 1 243, 1 81 does term ⎫ ⎬ begin to have integer values? ⎭ the sequence... 202. For which term does n − 1 ⎛ a n = − 36 ⎝ �
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�� ⎠ 2 3 first have a non-integer value? the geometric sequence 203. Use the recursive formula to write a geometric sequence whose common ratio is an integer. Show the first four terms, and then find the 10th term. Use the explicit 204. formula to write a geometric sequence whose common ratio is a decimal number between 0 and 1. Show the first 4 terms, and then find the 8th term. Is it possible for a sequence to be both arithmetic and 205. geometric? If so, give an example. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1487 13.4 | Series and Their Notations Learning Objectives 13.4.1 Use summation notation. 13.4.2 Use the formula for the sum of the first n terms of an arithmetic series. 13.4.3 Use the formula for the sum of the first n terms of a geometric series. 13.4.4 Use the formula for the sum of an infinite geometric series. 13.4.5 Solve annuity problems. A couple decides to start a college fund for their daughter. They plan to invest $50 in the fund each month. The fund pays 6% annual interest, compounded monthly. How much money will they have saved when their daughter is ready to start college in 6 years? In this section, we will learn how to answer this question. To do so, we need to consider the amount of money invested and the amount of interest earned. Using Summation Notation To find the total amount of money in the college fund and the sum of the amounts deposited, we need to add the amounts deposited each month and the amounts earned monthly. The sum of the terms of a sequence is called a series. Consider, for example, the following series. The nth partial sum of a series is the sum of a finite number of consecutive terms beginning with the first term. The notation Sn represents the partial sum. 3 + 7 + 11 + 15 + 19 +... S1 = 3 S2 = 3 + 7 = 10 S3 = 3 + 7 + 11 = 21 S4 = 3 + 7 + 11 + 15 = 36 Summation notation is used to represent series. Summation notation is often known as sigma notation because it uses the Greek capital letter sigma,
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Σ, to represent the sum. Summation notation includes an explicit formula and specifies the first and last terms in the series. An explicit formula for each term of the series is given to the right of the sigma. A variable called the index of summation is written below the sigma. The index of summation is set equal to the lower limit of summation, which is the number used to generate the first term in the series. The number above the sigma, called the upper limit of summation, is the number used to generate the last term in a series. If we interpret the given notation, we see that it asks us to find the sum of the terms in the series ak = 2k for k = 1 through k = 5. We can begin by substituting the terms for k and listing out the terms of this series. a1 = 2(1) = 2 a2 = 2(2) = 4 a3 = 2(3) = 6 a4 = 2(4) = 8 a5 = 2(5) = 10 We can find the sum of the series by adding the terms: 5 ∑ k = 1 2k = 2 + 4 + 6 + 8 + 10 = 30 1488 Chapter 13 Sequences, Probability, and Counting Theory Summation Notation The sum of the first n terms of a series can be expressed in summation notation as follows: n ∑ k = 1 ak This notation tells us to find the sum of ak from k = 1 to k = n. k is called the index of summation, 1 is the lower limit of summation, and n is the upper limit of summation. Does the lower limit of summation have to be 1? No. The lower limit of summation can be any number, but 1 is frequently used. We will look at examples with lower limits of summation other than 1. Given summation notation for a series, evaluate the value. 1. 2. Identify the lower limit of summation. Identify the upper limit of summation. 3. Substitute each value of k from the lower limit to the upper limit into the formula. 4. Add to find the sum. Example 13.21 Using Summation Notation 7 Evaluate ∑ k = 3 k 2. Solution According to the notation, the lower limit of summation is 3 and the upper limit is 7. So we need to find the sum of k 2 from k = 3 to k
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= 7. We find the terms of the series by substituting k = 3,4,5,6, and 7 into the function k 2. We add the terms to find the sum. 7 ∑ k = 3 k 2 = 32 + 42 + 52 + 62 + 72 = 9 + 16 + 25 + 36 + 49 = 135 13.25 5 Evaluate ∑ k = 2 (3k – 1). Using the Formula for Arithmetic Series Just as we studied special types of sequences, we will look at special types of series. Recall that an arithmetic sequence is a sequence in which the difference between any two consecutive terms is the common difference, d. The sum of the terms of an arithmetic sequence is called an arithmetic series. We can write the sum of the first n terms of an arithmetic series as: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1489 We can also reverse the order of the terms and write the sum as Sn = a1 + (a1 + d) + (a1 + 2d) +... + (an – d) + an. Sn = an + (an – d) + (an – 2d) +... + (a1 + d) + a1. If we add these two expressions for the sum of the first n terms of an arithmetic series, we can derive a formula for the sum of the first n terms of any arithmetic series. Sn = a1 + (a1 + d) + (a1 + 2d) +... + (an – d) + an + Sn = an + (an – d) + (an – 2d) +... + (a1 + d) + a1 2Sn = (a1 + an) + (a1 + an) +... + (a1 + an) Because there are n terms in the series, we can simplify this sum to 2Sn = n(a1 + an). We divide by 2 to find the formula for the sum of the first n terms of an arithmetic series. Sn = n(a1 + an) 2 Formula for the Sum of the First n Terms of an Arithmetic Series An arithmetic series is the sum of the terms of an arithmetic sequence. The formula for the sum of the first n terms of an arithmetic sequence is Sn = n(a1 + an) 2 (13.6)
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Given terms of an arithmetic series, find the sum of the first n terms. 1. Identify a1 and an. 2. Determine n. 3. Substitute values for a1, an, and n into the formula Sn = n(a1 + an) 2. 4. Simplify to find Sn. Example 13.22 Finding the First n Terms of an Arithmetic Series Find the sum of each arithmetic series. a. b. 5 + 8 + 11 + 14 + 17 + 20 + 23 + 26 + 29 + 32 20 + 15 + 10 +…+ −50 12 c. ∑ k = 1 3k − 8 Solution a. We are given a1 = 5 and an = 32. 1490 Chapter 13 Sequences, Probability, and Counting Theory Count the number of terms in the sequence to find n = 10. Substitute values for a1, an, and n into the formula and simplify. b. We are given a1 = 20 and an = − 50. S10 = Sn = n(a1 + an) 2 10(5 + 32) 2 = 185 Use the formula for the general term of an arithmetic sequence to find n. an = a1 + (n − 1)d −50 = 20 + (n − 1)( − 5) −70 = (n − 1)( − 5) 14 = n − 1 15 = n Substitute values for a1, an, n into the formula and simplify. Sn = n(a1 + an) 2 S15 = 15(20 − 50) 2 = − 225 c. To find a1, substitute k = 1 into the given explicit formula. ak = 3k − 8 a1 = 3(1) − 8 = − 5 We are given that n = 12. To find a12, substitute k = 12 into the given explicit formula. Substitute values for a1, an, and n into the formula and simplify. ak = 3k − 8 a12 = 3(12) − 8 = 28 Sn = n(a1 + an) 2 S12 = 12( − 5 + 28) 2 = 138 Use the formula to find the sum of each arithmetic series. 13.26 1.4 + 1.6 + 1.8 + 2.0 + 2.2 + 2.4 + 2.6 + 2.8 + 3.0 + 3.2 + 3.4 13.27 13 + 21 + 29 + … +
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69 13.28 10 ∑ k = 1 5 − 6k This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1491 Example 13.23 Solving Application Problems with Arithmetic Series On the Sunday after a minor surgery, a woman is able to walk a half-mile. Each Sunday, she walks an additional quarter-mile. After 8 weeks, what will be the total number of miles she has walked? Solution This problem can be modeled by an arithmetic series with a1 = 1 2 number of miles walked after 8 weeks, so we know that n = 8, and we are looking for S8. To find a8, we can use the explicit formula for an arithmetic sequence.. We are looking for the total and d = 1 4 an = a1 + d(n − 1) a8 = 1 + 1 2 4 (8 − 1) = 9 4 We can now use the formula for arithmetic series. Sn = S8 = She will have walked a total of 11 miles. n(a1 + an) 2 2 + 9 4) 2 8(1 = 11 A man earns $100 in the first week of June. Each week, he earns $12.50 more than the previous week. 13.29 After 12 weeks, how much has he earned? Using the Formula for Geometric Series Just as the sum of the terms of an arithmetic sequence is called an arithmetic series, the sum of the terms in a geometric sequence is called a geometric series. Recall that a geometric sequence is a sequence in which the ratio of any two consecutive terms is the common ratio, r. We can write the sum of the first n terms of a geometric series as Sn = a1 + ra1 + r 2 a1 +... + r n – 1 a1. Just as with arithmetic series, we can do some algebraic manipulation to derive a formula for the sum of the first n terms of a geometric series. We will begin by multiplying both sides of the equation by r. Next, we subtract this equation from the original equation. rSn = ra1 + r 2 a1 + r 3 a1 +... + r n a1 Sn = a1 + ra1 + r 2 a1 +... + r n – 1 a1 −rSn = − (ra1 + r 2 a1 + r 3 a1 +... + r n a1)
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(1 − r)Sn = a1 − r n a1 Notice that when we subtract, all but the first term of the top equation and the last term of the bottom equation cancel out. To obtain a formula for Sn, divide both sides by (1 − r). Sn = a1( 1492 Chapter 13 Sequences, Probability, and Counting Theory Formula for the Sum of the First n Terms of a Geometric Series A geometric series is the sum of the terms in a geometric sequence. The formula for the sum of the first n terms of a geometric sequence is represented as Sn = a1(13.7) Given a geometric series, find the sum of the first n terms. 1. Identify a1, r, and n. 2. Substitute values for a1, r, and n into the formula Sn = a1(. Simplify to find Sn. Example 13.24 Finding the First n Terms of a Geometric Series Use the formula to find the indicated partial sum of each geometric series. a. S11 for the series 8 + -4 + 2 + … k = 1 b. ∑ 6 3 ⋅ 2 k Solution a. a1 = 8, and we are given that n = 11. We can find r by dividing the second term of the series by the first. Substitute values for a1, r, r = −4 8 and n into the formula and simplify. = − 1 2 Sn = S11 = ) a1 (1 − r n 1 − r ⎛ ⎛ ⎝− 1 ⎝1 − 2 ⎛ ⎝− 1 1 − 2 8 ⎞ ⎠ ⎞ ⎠ 11⎞ ⎠ ≈ 5.336 b. Find a1 by substituting k = 1 into the given explicit formula. a1 = 3 ⋅ 21 = 6 We can see from the given explicit formula that r = 2. The upper limit of summation is 6, so n = 6. Substitute values for a1, r, and n into the formula, and simplify. Sn = S6 = a1(1 − r n 1 − r 6(1 − 26) 1 − 2 ) = 378 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1493 Use the formula to find the indicated
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partial sum of each geometric series. 13.30 S20 for the series 1,000 + 500 + 250 + … 13.31 8 ∑ k = 1 k 3 Example 13.25 Solving an Application Problem with a Geometric Series At a new job, an employee’s starting salary is $26,750. He receives a 1.6% annual raise. Find his total earnings at the end of 5 years. Solution The problem can be represented by a geometric series with a1 = 26, 750; n = 5; and r = 1.016. Substitute values for a1, r, and n into the formula and simplify to find the total amount earned at the end of 5 years. Sn = S5 = a1(1 − r n 1 − r ) 26,750(1 − 1.0165) 1 − 1.016 ≈ 138,099.03 He will have earned a total of $138,099.03 by the end of 5 years. At a new job, an employee’s starting salary is $32,100. She receives a 2% annual raise. How much will 13.32 she have earned by the end of 8 years? Using the Formula for the Sum of an Infinite Geometric Series Thus far, we have looked only at finite series. Sometimes, however, we are interested in the sum of the terms of an infinite sequence rather than the sum of only the first n terms. An infinite series is the sum of the terms of an infinite sequence. An example of an infinite series is 2 + 4 + 6 + 8 +... ∞ This series can also be written in summation notation as ∑ k = 1 2k, where the upper limit of summation is infinity. Because the terms are not tending to zero, the sum of the series increases without bound as we add more terms. Therefore, the sum of this infinite series is not defined. When the sum is not a real number, we say the series diverges. Determining Whether the Sum of an Infinite Geometric Series is Defined If the terms of an infinite geometric series approach 0, the sum of an infinite geometric series can be defined. The terms in this series approach 0: 1 + 0.2 + 0.04 + 0.008 + 0.0016 +... The common ratio r = 0.2. As n gets very large, the values of r n get very small and approach 0. Each successive term affects the sum less
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than the preceding term. As each succeeding term gets closer to 0, the sum of the terms approaches a finite value. The terms of any infinite geometric series with −1 < r < 1 approach 0; the sum of a geometric series is defined when −1 < r < 1. 1494 Chapter 13 Sequences, Probability, and Counting Theory Determining Whether the Sum of an Infinite Geometric Series is Defined The sum of an infinite series is defined if the series is geometric and −1 < r < 1. Given the first several terms of an infinite series, determine if the sum of the series exists. 1. Find the ratio of the second term to the first term. 2. Find the ratio of the third term to the second term. 3. Continue this process to ensure the ratio of a term to the preceding term is constant throughout. If so, the series is geometric. 4. If a common ratio, r, was found in step 3, check to see if −1 < r < 1. If so, the sum is defined. If not, the sum is not defined. Example 13.26 Determining Whether the Sum of an Infinite Series is Defined Determine whether the sum of each infinite series is defined. a. b. 12 + +... ∞ c. ∑ k = 1 k 27 ⋅ (1 3 ) ∞ d. ∑ k = 1 5k Solution a. The ratio of the second term to the first is 2 3 second, 1 2. The series is not geometric., which is not the same as the ratio of the third term to the b. The ratio of the second term to the first is the same as the ratio of the third term to the second. The series is geometric with a common ratio of 2 3. The sum of the infinite series is defined. c. The given formula is exponential with a base of 1 3 ; the series is geometric with a common ratio of 1 3. The sum of the infinite series is defined. d. The given formula is not exponential; the series is not geometric because the terms are increasing, and so cannot yield a finite sum. Determine whether the sum of the infinite series is defined. 13.33 +... This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1495 13.34 24 + (−12) + 6 + (−3) +...
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13.35 ∞ ∑ k = 1 15 ⋅ ( – 0.3) k Finding Sums of Infinite Series When the sum of an infinite geometric series exists, we can calculate the sum. The formula for the sum of an infinite series is related to the formula for the sum of the first n terms of a geometric series. We will examine an infinite series with r = 1 2. What happens to r n as n increases? Sn = a1( 16 The value of r n decreases rapidly. What happens for greater values of n? 10 20 30 (1 2 ) (1 2 ) (1 2 ) = 1 1,024 = = 1 1,048,576 1 1,073,741,824 As n gets very large, r n gets very small. We say that, as n increases without bound, r n approaches 0. As r n approaches 0, 1 − r n approaches 1. When this happens, the numerator approaches a1. This give us a formula for the sum of an infinite geometric series. Formula for the Sum of an Infinite Geometric Series The formula for the sum of an infinite geometric series with −1 < r < 1 is S = a1 1 − r (13.8) Given an infinite geometric series, find its sum. 1. Identify a1 and r. 2. Confirm that – 1 < r < 1. 3. Substitute values for a1 and r into the formula, S = a1 1 − r. 4. Simplify to find S. 1496 Chapter 13 Sequences, Probability, and Counting Theory Example 13.27 Finding the Sum of an Infinite Geometric Series Find the sum, if it exists, for the following: a. b. 10 + 9 + 8 + 7 + … 248.6 + 99.44 + 39.776 + … ∞ c. ∑ k = 1 k – 1 4,374 ⋅ ( – 1 3 ) ∞ d4 3 ) Solution a. There is not a constant ratio; the series is not geometric. b. There is a constant ratio; the series is geometric. a1 = 248.6 and r = 99.44 248.6 = 0.4, Substitute a1 = 248.6 and r = 0.4 into the formula and simplify to find the sum: so the sum exists. S = a1 1 − r S = 248.6 1 − 0.4 ¯ = 414.
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3 c. The formula is exponential, so the series is geometric with r = – 1 3. Find a1 by substituting k = 1 into the given explicit formula: a1 = 4,374 ⋅ ( – 1 3 ) 1 – 1 = 4,374 Substitute a1 = 4,374 and r = − 1 3 into the formula, and simplify to find the sum: S = a1 1 − r S = 4,374 1 − ( − 1 3) = 3,280.5 d. The formula is exponential, so the series is geometric, but r > 1. The sum does not exist. Example 13.28 Finding an Equivalent Fraction for a Repeating Decimal ¯ Find an equivalent fraction for the repeating decimal 0. 3 Solution ¯ We notice the repeating decimal 0. 3 = 0.333... so we can rewrite the repeating decimal as a sum of terms. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1497 Looking for a pattern, we rewrite the sum, noticing that we see the first term multiplied to 0.1 in the second term, and the second term multiplied to 0.1 in the third term. ¯ 0. 3 = 0.3 + 0.03 + 0.003 +... Notice the pattern; we multiply each consecutive term by a common ratio of 0.1 starting with the first term of 0.3. So, substituting into our formula for an infinite geometric sum, we have Sn = a1 1 − r = 0.3 1 − 0.1 = 0.3 0.9 = 1 3. Find the sum, if it exists. 13.36 2 + 2 3 + 2 9 +... 13.37 ∞ ∑ k = 1 0.76k + 1 13.38 ∞ ∑ k = 1 k ⎛ ⎝− 3 8 ⎞ ⎠ Solving Annuity Problems At the beginning of the section, we looked at a problem in which a couple invested a set amount of money each month into a college fund for six years. An annuity is an investment in which the purchaser makes a sequence of periodic, equal payments. To find the amount of an annuity, we need to find the sum of all the payments and the interest earned. In the example, the couple invests $50 each month. This is the value of the
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initial deposit. The account paid 6% annual interest, compounded monthly. To find the interest rate per payment period, we need to divide the 6% annual percentage interest (APR) rate by 12. So the monthly interest rate is 0.5%. We can multiply the amount in the account each month by 100.5% to find the value of the account after interest has been added. We can find the value of the annuity right after the last deposit by using a geometric series with a1 = 50 and r = 100.5% = 1.005. After the first deposit, the value of the annuity will be $50. Let us see if we can determine the amount in the college fund and the interest earned. We can find the value of the annuity after n deposits using the formula for the sum of the first n terms of a geometric series. In 6 years, there are 72 months, so n = 72. We can substitute a1 = 50, and n = 72 into the formula, and simplify to find the value of the annuity after 6 years. r = 1.005, S72 = 50(1 − 1.00572) 1 − 1.005 ≈ 4,320.44 After the last deposit, the couple will have a total of $4,320.44 in the account. Notice, the couple made 72 payments of $50 each for a total of 72(50) = $3,600. This means that because of the annuity, the couple earned $720.44 interest in their college fund. 1498 Chapter 13 Sequences, Probability, and Counting Theory Given an initial deposit and an interest rate, find the value of an annuity. 1. Determine a1, the value of the initial deposit. 2. Determine n, the number of deposits. 3. Determine r. a. Divide the annual interest rate by the number of times per year that interest is compounded. b. Add 1 to this amount to find r. 4. Substitute values for a1, r, and n into the formula for the sum of the first n terms of a geometric series, Sn = a1(. Simplify to find Sn, the value of the annuity after n deposits. Example 13.29 Solving an Annuity Problem A deposit of $100 is placed into a college fund at the beginning of every month for 10 years. The fund earns 9% annual interest, compounded monthly, and paid at the end of
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the month. How much is in the account right after the last deposit? Solution The value of the initial deposit is $100, so a1 = 100. A total of 120 monthly deposits are made in the 10 years, so n = 120. To find r, divide the annual interest rate by 12 to find the monthly interest rate and add 1 to represent the new monthly deposit. r = 1 + 0.09 12 Substitute a1 = 100, r = 1.0075, and n = 120 into the formula for the sum of the first n terms of a geometric series, and simplify to find the value of the annuity. = 1.0075 S120 = 100(1 − 1.0075120) 1 − 1.0075 ≈ 19,351.43 So the account has $19,351.43 after the last deposit is made. 13.39 At the beginning of each month, $200 is deposited into a retirement fund. The fund earns 6% annual interest, compounded monthly, and paid into the account at the end of the month. How much is in the account if deposits are made for 10 years? Access these online resources for additional instruction and practice with series. • Arithmetic Series (http://openstaxcollege.org/l/arithmeticser) • Geometric Series (http://openstaxcollege.org/l/geometricser) • Summation Notation (http://openstaxcollege.org/l/sumnotation) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1499 13.4 EXERCISES Verbal 206. What is an nth partial sum? For the following exercises, use the formula for the sum of the first n terms of each geometric sequence, and then state the indicated sum. What 207. sequence and an arithmetic series? is the difference between an arithmetic 208. What is a geometric series? How is finding the sum of an infinite geometric series 209. different from finding the nth partial sum? 210. What is an annuity? Algebraic For the following exercises, express each description of a sum using summation notation. 211. The sum of terms m2 + 3m from m = 1 to m = 5 212. The sum from of n = 0 to n = 4 of 5n 213. The sum of 6k − 5 from k
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= − 2 to k = 1 The sum that results from adding the number 4 five 214. times For the following exercises, express each arithmetic sum using summation notation. 224. 225. 226 11 ∑ a = 1 n − 1 5 ⋅ 2 64 ⋅ 0.2 a − 1 For the following exercises, determine whether the infinite series has a sum. If so, write the formula for the sum. If not, state the reason. 227. 228. 229. 230. 12 + 18 + 24 + 30 +... 2 + 1.6 + 1.28 + 1.024 +... ∞ ∑ − ⎛ ⎝− 1 2 ⎞ ⎠ 215. 216. 217. 5 + 10 + 15 + 20 + 25 + 30 + 35 + 40 + 45 + 50 Graphical 10 + 18 + 26 + … + 162 For the following exercises, use the following scenario. Javier makes monthly deposits into a savings account. He opened the account with an initial deposit of $50. Each month thereafter he increased the previous deposit amount by $20. For the following exercises, use the formula for the sum of the first n terms of each arithmetic sequence. Graph the arithmetic sequence showing one year of 231. Javier’s deposits. 218. 219. 220 Graph the arithmetic series showing the monthly 232. sums of one year of Javier’s deposits. 19 + 25 + 31 + … + 73 For the following exercises, use the geometric series 3.2 + 3.4 + 3.6 + … + 5. For the following exercises, express each geometric sum using summation notation. 233. Graph the first 7 partial sums of the series. 221. 222. 223. 1 + 3 + 9 + 27 + 81 + 243 + 729 + 2187 234. What number does Sn seem to be approaching in the 8 + 4 + 2 + … + 0.125 − 1 6 + 1 12 − 1 24 + … + 1 768 graph? Find the sum to explain why this makes sense. Numeric For the following exercises, find the indicated sum. 235. 1500 14 ∑ a = 1 a 236. 237. 238. 6 ∑ n = 1 n(n − 2) 17 ∑ For the following exercises, use the formula for the sum of the first n terms of an arithmetic series to find the sum. 239. 240. 241. 242. −1.7 + − 0.4 + 0.9 + 2
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.2 + 3.5 + 4.8 6 + 15 2 + 9 + 21 2 + 12 + 27 2 + 15 −1 + 3 + 7 +... + 31 11 ∑ For the following exercises, use the formula for the sum of the first n terms of a geometric series to find the partial sum. 243. 244. 245. 246. S6 S7 for the series −2 − 10 − 50 − 250... for the series 0.4 − 2 + 10 − 50... 9 ∑ k = 1 10 ∑ 2 ⋅ ⎞ ⎠ ⎛ ⎝ 1 2 For the following exercises, find the sum of the infinite geometric series. 4 + 2 + 1 + 1 2... −1 − 1 4 − 1 16 − 1 64... 247. 248. 249. 250. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory ∞ ∑ n = 1 4.6 ⋅ 0.5 n − 1 For the following exercises, determine the value of the annuity for the indicated monthly deposit amount, the number of deposits, and the interest rate. 251. Deposit amount: $50; rate: 5%, compounded monthly total deposits: 60; interest Deposit amount: $150; total deposits: 24; interest 252. rate: 3%, compounded monthly Deposit amount: $450; total deposits: 60; interest 253. rate: 4.5%, compounded quarterly Deposit amount: $100; total deposits: 120; interest 254. rate: 10%, compounded semi-annually Extensions 255. The sum of terms 50 − k 2 from k = x through 7 is 115. What is x? 256. Write an explicit formula for ak such that 6 ∑ k = 0 257. n ∑ k = 1 ak = 189. Assume this is an arithmetic series. Find the smallest value of n such that (3k – 5) > 100. How many terms must be added before the series 258. −1 − 3 − 5 − 7.... has a sum less than −75? 259. Write 0.65 as an infinite geometric series using summation notation. Then use the formula for finding the sum of an infinite geometric series to convert 0.65 to a fraction. 260. The sum of an infinite geometric series is five times the value of the first term. What is the common ratio of the series?
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261. To get the best loan rates available, the Riches want to save enough money to place 20% down on a $160,000 home. They plan to make monthly deposits of $125 in an investment account interest compounded semi-annually. Will the Riches have enough for a 20% down payment after five years of saving? How much money will they have saved? that offers 8.5% annual 262. Chapter 13 Sequences, Probability, and Counting Theory 1501 Karl has two years to save $10, 000 to buy a used car when he graduates. To the nearest dollar, what would his monthly deposits need to be if he invests in an account offering a 4.2% annual rate that compounds monthly? interest Real-World Applications Keisha devised a week-long study plan to prepare for 263. finals. On the first day, she plans to study for 1 hour, and each successive day she will increase her study time by 30 minutes. How many hours will Keisha have studied after one week? A boulder rolled down a mountain, traveling 6 feet in 264. the first second. Each successive second, its distance increased by 8 feet. How far did the boulder travel after 10 seconds? A scientist places 50 cells in a petri dish. Every hour, 265. the population increases by 1.5%. What will the cell count be after 1 day? A pendulum travels a distance of 3 feet on its first 266. swing. On each successive swing, it travels 3 4 the distance of the previous swing. What is the total distance traveled by the pendulum when it stops swinging? Rachael deposits $1,500 into a retirement fund each 267. year. The fund earns 8.2% annual interest, compounded monthly. If she opened her account when she was 19 years old, how much will she have by the time she is 55? How much of that amount will be interest earned? 1502 Chapter 13 Sequences, Probability, and Counting Theory 13.5 | Counting Principles Learning Objectives In this section, you will: 13.5.1 Solve counting problems using the Addition Principle. 13.5.2 Solve counting problems using the Multiplication Principle. 13.5.3 Solve counting problems using permutations involving n distinct objects. 13.5.4 Solve counting problems using combinations. 13.5.5 Find the number of subsets of a given set. 13.5.6 Solve counting problems using perm
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utations involving n non-distinct objects. A new company sells customizable cases for tablets and smartphones. Each case comes in a variety of colors and can be personalized for an additional fee with images or a monogram. A customer can choose not to personalize or could choose to have one, two, or three images or a monogram. The customer can choose the order of the images and the letters in the monogram. The company is working with an agency to develop a marketing campaign with a focus on the huge number of options they offer. Counting the possibilities is challenging! We encounter a wide variety of counting problems every day. There is a branch of mathematics devoted to the study of counting problems such as this one. Other applications of counting include secure passwords, horse racing outcomes, and college scheduling choices. We will examine this type of mathematics in this section. Using the Addition Principle The company that sells customizable cases offers cases for tablets and smartphones. There are 3 supported tablet models and 5 supported smartphone models. The Addition Principle tells us that we can add the number of tablet options to the number of smartphone options to find the total number of options. By the Addition Principle, there are 8 total options, as we can see in Figure 13.18. Figure 13.18 The Addition Principle According to the Addition Principle, if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. Example 13.30 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1503 Using the Addition Principle There are 2 vegetarian entrée options and 5 meat entrée options on a dinner menu. What is the total number of entrée options? Solution We can add the number of vegetarian options to the number of meat options to find the total number of entrée options. There are 7 total options. A student is shopping for a new computer. He is deciding among 3 desktop computers and 4 laptop 13.40 computers. What is the total number of computer options? Using the Multiplication Principle The Multiplication Principle applies when we are making more than one selection. Suppose we are choosing an appetizer, an entrée, and a dessert. If there are 2 appetizer options, 3 entrée options, and 2 dessert options on a fixed-price dinner menu, there are a total of 12
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possible choices of one each as shown in the tree diagram in Figure 13.19. Figure 13.19 The possible choices are: 1. 2. 3. 4. 5. 6. 7. 8. 9. soup, chicken, cake soup, chicken, pudding soup, fish, cake soup, fish, pudding soup, steak, cake soup, steak, pudding salad, chicken, cake salad, chicken, pudding salad, fish, cake 10. salad, fish, pudding 1504 Chapter 13 Sequences, Probability, and Counting Theory 11. salad, steak, cake 12. salad, steak, pudding We can also find the total number of possible dinners by multiplying. We could also conclude that there are 12 possible dinner choices simply by applying the Multiplication Principle. # of appetizer options × # of entree options × # of dessert options 2 × 3 × 2 = 12 The Multiplication Principle According to the Multiplication Principle, if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways. This is also known as the Fundamental Counting Principle. Example 13.31 Using the Multiplication Principle Diane packed 2 skirts, 4 blouses, and a sweater for her business trip. She will need to choose a skirt and a blouse for each outfit and decide whether to wear the sweater. Use the Multiplication Principle to find the total number of possible outfits. Solution To find the total number of outfits, find the product of the number of skirt options, the number of blouse options, and the number of sweater options. There are 16 possible outfits. 13.41 A restaurant offers a breakfast special that includes a breakfast sandwich, a side dish, and a beverage. There are 3 types of breakfast sandwiches, 4 side dish options, and 5 beverage choices. Find the total number of possible breakfast specials. Finding the Number of Permutations of n Distinct Objects The Multiplication Principle can be used to solve a variety of problem types. One type of problem involves placing objects in order. We arrange letters into words and digits into numbers, line up for photographs, decorate rooms, and more. An ordering of objects is called a permutation. Finding the Number of Permutations of n Distinct Objects Using the Multiplication Principle To solve permutation problems, it is often helpful to draw line segments for each option. That enables us to determine the number of each option so
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we can multiply. For instance, suppose we have four paintings, and we want to find the number of ways we can hang three of the paintings in order on the wall. We can draw three lines to represent the three places on the wall. There are four options for the first place, so we write a 4 on the first line. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1505 After the first place has been filled, there are three options for the second place so we write a 3 on the second line. After the second place has been filled, there are two options for the third place so we write a 2 on the third line. Finally, we find the product. There are 24 possible permutations of the paintings. Given n distinct options, determine how many permutations there are. 1. Determine how many options there are for the first situation. 2. Determine how many options are left for the second situation. 3. Continue until all of the spots are filled. 4. Multiply the numbers together. Example 13.32 Finding the Number of Permutations Using the Multiplication Principle At a swimming competition, nine swimmers compete in a race. a. How many ways can they place first, second, and third? b. How many ways can they place first, second, and third if a swimmer named Ariel wins first place? (Assume there is only one contestant named Ariel.) c. How many ways can all nine swimmers line up for a photo? Solution a. Draw lines for each place. There are 9 options for first place. Once someone has won first place, there are 8 remaining options for second place. Once first and second place have been won, there are 7 remaining options for third place. Multiply to find that there are 504 ways for the swimmers to place. b. Draw lines for describing each place. 1506 Chapter 13 Sequences, Probability, and Counting Theory We know Ariel must win first place, so there is only 1 option for first place. There are 8 remaining options for second place, and then 7 remaining options for third place. Multiply to find that there are 56 ways for the swimmers to place if Ariel wins first. c. Draw lines for describing each place in the photo. There are 9 choices for the first spot, then 8 for the second, 7 for the third,
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6 for the fourth, and so on until only 1 person remains for the last spot. There are 362,880 possible permutations for the swimmers to line up. Analysis Note that in part c, we found there were 9! ways for 9 people to line up. The number of permutations of n distinct objects can always be found by n!. A family of five is having portraits taken. Use the Multiplication Principle to find the following. 13.42 How many ways can the family line up for the portrait? 13.43 How many ways can the photographer line up 3 family members? 13.44 How many ways can the family line up for the portrait if the parents are required to stand on each end? Finding the Number of Permutations of n Distinct Objects Using a Formula For some permutation problems, it is inconvenient to use the Multiplication Principle because there are so many numbers to multiply. Fortunately, we can solve these problems using a formula. Before we learn the formula, let’s look at two common notations for permutations. If we have a set of n objects and we want to choose r objects from the set in order, we write P(n, r). Another way to write this is nPr, a notation commonly seen on computers and calculators. To calculate P(n, r), we begin by finding n!, the number of ways to line up all n objects. We then divide by (n − r)! to cancel out the (n − r) items that we do not wish to line up. Let’s see how this works with a simple example. Imagine a club of six people. They need to elect a president, a vice president, and a treasurer. Six people can be elected president, any one of the five remaining people can be elected vice president, and any of the remaining four people could be elected treasurer. The number of ways this may be done is 6×5×4 = 120. Using factorials, we get the same result. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1507 There are 120 ways to select 3 officers in order from a club with 6 members. We refer to this as a permutation of 6 taken 3 at a time. The general formula is as follows = 120 P(n, r) = n! (n − r)! Note
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that the formula stills works if we are choosing all n objects and placing them in order. In that case we would be dividing by (n − n)! or 0!, which we said earlier is equal to 1. So the number of permutations of n objects taken n at a time is n! 1 or just n!. Formula for Permutations of n Distinct Objects Given n distinct objects, the number of ways to select r objects from the set in order is P(n, r) = n! (n − r)! (13.9) Given a word problem, evaluate the possible permutations. 1. 2. Identify n from the given information. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 13.33 Finding the Number of Permutations Using the Formula A professor is creating an exam of 9 questions from a test bank of 12 questions. How many ways can she select and arrange the questions? Solution Substitute n = 12 and r = 9 into the permutation formula and simplify. P(n, r) = P(12, 9) = n! (n − r)! 12! (12 − 9)! = 12! 3! = 79,833,600 There are 79,833,600 possible permutations of exam questions! Analysis We can also use a calculator to find permutations. For this problem, we would enter 15, press the n Pr function, enter 12, and then press the equal sign. The n Pr function may be located under the MATH menu with probability commands. Could we have solved Example 13.33 using the Multiplication Principle? Yes. We could have multiplied 15 ⋅ 14 ⋅ 13 ⋅ 12 ⋅ 11 ⋅ 10 ⋅ to find the same answer. 1508 Chapter 13 Sequences, Probability, and Counting Theory A play has a cast of 7 actors preparing to make their curtain call. Use the permutation formula to find the following. 13.45 How many ways can the 7 actors line up? 13.46 How many ways can 5 of the 7 actors be chosen to line up? Find the Number of Combinations Using the Formula So far, we have looked at problems asking us to put objects in order. There are many problems in which we want to select a few objects from a group of objects, but we do not care about the order. When
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we are selecting objects and the order does not matter, we are dealing with combinations. A selection of r objects from a set of n objects where the order does not matter can be written as C(n, r). Just as with permutations, C(n, r) can also be written as n Cr. In this case, the general formula is as follows. C(n, r) = n! r!(n − r)! An earlier problem considered choosing 3 of 4 possible paintings to hang on a wall. We found that there were 24 ways to select 3 of the 4 paintings in order. But what if we did not care about the order? We would expect a smaller number because selecting paintings 1, 2, 3 would be the same as selecting paintings 2, 3, 1. To find the number of ways to select 3 of the 4 paintings, disregarding the order of the paintings, divide the number of permutations by the number of ways to order 3 paintings. There are ways to order 3 paintings. There are 24, or 4 ways to select 3 of the 4 paintings. 6 This number makes sense because every time we are selecting 3 paintings, we are not selecting 1 painting. There are 4 paintings we could choose not to select, so there are 4 ways to select 3 of the 4 paintings. Formula for Combinations of n Distinct Objects Given n distinct objects, the number of ways to select r objects from the set is C(n, r) = n! r!(n − r)! (13.10) Given a number of options, determine the possible number of combinations. 1. 2. Identify n from the given information. Identify r from the given information. 3. Replace n and r in the formula with the given values. 4. Evaluate. Example 13.34 Finding the Number of Combinations Using the Formula A fast food restaurant offers five side dish options. Your meal comes with two side dishes. a. How many ways can you select your side dishes? b. How many ways can you select 3 side dishes? This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1509 Solution a. We want to choose 2 side dishes from 5 options. b. We want to choose 3 side dishes from 5 options. C(5, 2) = 5! 2!(5 − 2)! = 10 C(5,
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3) = 5! 3!(5 − 3)! = 10 Analysis We can also use a graphing calculator to find combinations. Enter 5, then press n Cr, enter 3, and then press the equal sign. The n Cr, function may be located under the MATH menu with probability commands. Is it a coincidence that parts (a) and (b) in Example 13.34 have the same answers? No. When we choose r objects from n objects, we are not choosing (n – r) objects. Therefore, C(n, r) = C(n, n – r). An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a 13.47 banana split? Finding the Number of Subsets of a Set We have looked only at combination problems in which we chose exactly r objects. In some problems, we want to consider choosing every possible number of objects. Consider, for example, a pizza restaurant that offers 5 toppings. Any number of toppings can be ordered. How many different pizzas are possible? To answer this question, we need to consider pizzas with any number of toppings. There is C(5, 0) = 1 way to order a pizza with no toppings. There are C(5, 1) = 5 ways to order a pizza with exactly one topping. If we continue this process, we get C(5, 0) + C(5, 1) + C(5, 2) + C(5, 3) + C(5, 4) + C(5, 5) = 32 There are 32 possible pizzas. This result is equal to 25. We are presented with a sequence of choices. For each of the n objects we have two choices: include it in the subset or not. So for the whole subset we have made n choices, each with two options. So there are a total of 2 · 2 · 2 · … · 2 possible resulting subsets, all the way from the empty subset, which we obtain when we say “no” each time, to the original set itself, which we obtain when we say “yes” each time. Formula for the Number of Subsets of a Set A set containing n distinct objects has 2 n subsets. Example 13.35 Finding the Number of Subsets of a Set 1510 Chapter 13 Sequences, Probability, and Counting Theory A restaurant offers butter, cheese, chives
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, and sour cream as toppings for a baked potato. How many different ways are there to order a potato? Solution We are looking for the number of subsets of a set with 4 objects. Substitute n = 4 into the formula. There are 16 possible ways to order a potato. n 2 = 24 = 16 A sundae bar at a wedding has 6 toppings to choose from. Any number of toppings can be chosen. How 13.48 many different sundaes are possible? Finding the Number of Permutations of n Non-Distinct Objects We have studied permutations where all of the objects involved were distinct. What happens if some of the objects are indistinguishable? For example, suppose there is a sheet of 12 stickers. If all of the stickers were distinct, there would be 12! ways to order the stickers. However, 4 of the stickers are identical stars, and 3 are identical moons. Because all of the objects are not distinct, many of the 12! permutations we counted are duplicates. The general formula for this situation is as follows. n! r1!r2! … rk! In this example, we need to divide by the number of ways to order the 4 stars and the ways to order the 3 moons to find the number of unique permutations of the stickers. There are 4! ways to order the stars and 3! ways to order the moon. There are 3,326,400 ways to order the sheet of stickers. 12! 4!3! = 3,326,400 Formula for Finding the Number of Permutations of n Non-Distinct Objects If there are n elements in a set and r1 are alike, r2 are alike, r3 are alike, and so on through rk, the number of permutations can be found by n! r1!r2! … rk! (13.11) Example 13.36 Finding the Number of Permutations of n Non-Distinct Objects Find the number of rearrangements of the letters in the word DISTINCT. Solution There are 8 letters. Both I and T are repeated 2 times. Substitute n = 8, formula. r1 = 2, and r2 = 2 into the This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1511 There are 10,080 arrangements. 8! 2!2! = 10,080
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13.49 Find the number of rearrangements of the letters in the word CARRIER. Access these online resources for additional instruction and practice with combinations and permutations. • Combinations (http://openstaxcollege.org/l/combinations) • Permutations (http://openstaxcollege.org/l/permutations) 1512 Chapter 13 Sequences, Probability, and Counting Theory 13.5 EXERCISES Verbal For the following exercises, assume that there are n ways an event A can happen, m ways an event B can happen, and that A and B are non-overlapping. Use the Addition Principle of counting to explain how 268. many ways event A or B can occur. Use the Multiplication Principle of counting to 269. explain how many ways event A and B can occur. Answer the following questions. How many two-letter strings—the first letter from A 279. and the second letter from B — can be formed from the sets A = {b, c, d} and B = {a, e, i, o, u}? How many ways are there to construct a string of 3 280. digits if numbers can be repeated? How many ways are there to construct a string of 3 281. digits if numbers cannot be repeated? For the following exercises, compute the value of the expression. to apply 270. When given two separate events, how do we know or the whether Multiplication possible outcomes? What conjunctions may help to determine which operations to use? the Addition Principle calculating Principle when 271. Describe how the permutation of n objects differs from the permutation of choosing r objects from a set of n objects. Include how each is calculated. 272. What is the term for the arrangement that selects r objects from a set of n objects when the order of the r objects is not is the formula for calculating the number of possible outcomes for this type of arrangement? important? What Numeric For the following exercises, determine whether to use the Addition Principle or the Multiplication Principle. Then perform the calculations. Let 273. the set A = { − 5, − 3, − 1, 2, 3, 4, 5, 6}. How many ways are there to choose a negative or an even number from A? the Let 274. B = { − 23, − 16, − 7, − 2, 20, 36, 48, 72}. many ways are there to choose a positive or
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an odd number from A? set How How many ways are there to pick a red ace or a club 275. from a standard card playing deck? How many ways are there to pick a paint color from 5 276. shades of green, 4 shades of blue, or 7 shades of yellow? How many outcomes are possible from tossing a pair 277. of coins? How many outcomes are possible from tossing a coin 278. and rolling a 6-sided die? This content is available for free at https://cnx.org/content/col11758/1.5 282. P(5, 2) 283. P(8, 4) 284. P(3, 3) 285. P(9, 6) 286. P(11, 5) 287. C(8, 5) 288. C(12, 4) 289. C(26, 3) 290. C(7, 6) 291. C(10, 3) For the following exercises, find the number of subsets in each given set. 292. {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} 293. {a, b, c, …, z} A set containing 5 distinct numbers, 4 distinct letters, 294. and 3 distinct symbols 295. The set of even numbers from 2 to 28 The set of two-digit numbers between 1 and 100 296. containing the digit 0 For the following exercises, find the distinct number of arrangements. 297. The letters in the word “juggernaut” 298. The letters in the word “academia” 299. Chapter 13 Sequences, Probability, and Counting Theory 1513 The letters in the word “academia” that begin and end in “a” 300. The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% The symbols in the string #,#,#,@,@,$,$,$,%,%,%,% 301. that begin and end with “%” Extensions 302. The set, S consists of 900,000,000 whole numbers, each being the same number of digits long. How many digits long is a number from S? (Hint: use the fact that a whole number cannot start with the digit 0.) subsets The number of 5-element from a set 303. containing n elements is equal to the number of 6-element subsets from the
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same set. What is the value of n? (Hint: the order in which the elements for the subsets are chosen is not important.) 304. Can C(n, r) ever equal P(n, r)? Explain. Suppose a set A has 2,048 subsets. How many 305. distinct objects are contained in A? Hector wants to place billboard advertisements throughout the county for his new business. How many ways can Hector choose 15 neighborhoods to advertise in if there are 30 neighborhoods in the county? An art store has 4 brands of paint pens in 12 different 312. colors and 3 types of ink. How many paint pens are there to choose from? How many ways can a committee of 3 freshmen and 4 313. juniors be formed from a group of 8 freshmen and 11 juniors? How many ways can a baseball coach arrange the 314. order of 9 batters if there are 15 players on the team? A conductor needs 5 cellists and 5 violinists to play at 315. a diplomatic event. To do this, he ranks the orchestra’s 10 cellists and 16 violinists in order of musical proficiency. What is the ratio of the total cellist rankings possible to the total violinist rankings possible? A motorcycle shop has 10 choppers, 6 bobbers, and 5 316. café racers—different types of vintage motorcycles. How many ways can the shop choose 3 choppers, 5 bobbers, and 2 café racers for a weekend showcase? 306. How many arrangements can be made from the letters of the word “mountains” if all the vowels must form a string? A skateboard shop stocks 10 types of board decks, 3 317. types of trucks, and 4 types of wheels. How many different skateboards can be constructed? Just-For-Kicks Sneaker Company offers an online 318. customizing service. How many ways are there to design a custom pair of Just-For-Kicks sneakers if a customer can choose from a basic shoe up to 11 customizable options? A car wash offers the following optional services to 319. the basic wash: clear coat wax, triple foam polish, undercarriage wash, rust inhibitor, wheel brightener, air freshener, and interior shampoo. How many washes are possible if any number of options can be added to the basic wash? Susan bought 20 plants to arrange along the border of 320. her garden. How many distinct arrangements can she make if the plants are comprised of 6 tulips,
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6 roses, and 8 daisies? How many unique ways can a string of Christmas 321. lights be arranged from 9 red, 10 green, 6 white, and 12 gold color bulbs? Real-World Applications A family consisting of 2 parents and 3 children is to 307. pose for a picture with 2 family members in the front and 3 in the back. a. How many arrangements are possible with no restrictions? b. How many arrangements are possible if the parents must sit in the front? c. How many arrangements are possible if the parents must be next to each other? A cell phone company offers 6 different voice 308. packages and 8 different data packages. Of those, 3 packages include both voice and data. How many ways are there to choose either voice or data, but not both? In horse racing, a “trifecta” occurs when a bettor wins 309. by selecting the first three finishers in the exact order (1st place, 2nd place, and 3rd place). How many different trifectas are possible if there are 14 horses in a race? A wholesale T-shirt company offers sizes small, 310. medium, large, and extra-large in organic or non-organic cotton and colors white, black, gray, blue, and red. How many different T-shirts are there to choose from? 311. 1514 Chapter 13 Sequences, Probability, and Counting Theory 13.6 | Binomial Theorem Learning Objectives In this section, you will: 13.6.1 Apply the Binomial Theorem. A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find (x + y) n without multiplying the binomial by itself n times. Identifying Binomial Coefficients In Counting Principles, we studied combinations. In the shortcut to finding (x + y) n to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation ⎛ r ⎝, we will need to use combinations ⎞ ⎠ instead of n C(n, r), but it can be calculated in the same way. So n () ⎛ r ⎝ ⎞ ⎠ = C(n
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, r) = n! r!(n − r)! n The combination ⎛ r ⎝ ⎞ ⎠ is called a binomial coefficient. An example of a binomial coefficient is ⎛ 5 ⎝ 2 ⎞ ⎠ = C(5, 2) = 10. Binomial Coefficients If n and r are integers greater than or equal to 0 with n ≥ r, then the binomial coefficient is n ⎛ r ⎝ ⎞ ⎠ = C(n, r) = n! r!(n − r)! Is a binomial coefficient always a whole number? Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number. Example 13.37 Finding Binomial Coefficients Find each binomial coefficient. a. b. c Solution This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1515 Use the formula to calculate each binomial coefficient. You can also use the nCr function on your calculator. n ⎛ r ⎝ ⎞ ⎠ = C(n, r) = n! r!(n − r)! a. b. c!(5 − 32! = 10 9! 2!(9 − 27! = 36 9! 7!(9 − 72! = 36 Analysis Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations. n () ⎛ ⎝ ⎞ ⎠ 13.50 Find each binomial coefficient. a. b. ⎞ ⎠ ⎛ 7 ⎝ 3 ⎛ 11 ⎝ 4 ⎞ ⎠ Using the Binomial Theorem n by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. When we expand (x + y) If we wanted to expand (x + y)52, we might multiply (x + y) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut
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for finding more complicated binomial expansions. (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2 y + 3xy2 + y3 (x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4 First, let’s examine the exponents. With each successive term, the exponent for x decreases and the exponent for y increases. The sum of the two exponents is n for each term. Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern: n ⎛ ⎝ 0 ⎞ ⎠, n ⎛ ⎝ 1 ⎞ ⎠, n ⎛ ⎝ 2 ⎞ ⎠,..., n ⎛ n ⎝ ⎞ ⎠. These patterns lead us to the Binomial Theorem, which can be used to expand any binomial. (x + y) n n = ∑ k = 0 = xn + n ⎛ ⎞ ⎠xn − kyk ⎝ k n ⎛ ⎝ 1 ⎞ ⎠xn − 1 y + n ⎛ ⎝ 2 ⎞ ⎠xn − 2 y2 +... + n ⎛ ⎞ ⎠xyn − 1 + yn ⎝ n − 1 1516 Chapter 13 Sequences, Probability, and Counting Theory Another way to see the coefficients is to examine the expansion of a binomial in general form, x + y, to successive powers 1, 2, 3, and 4. (x + y)1 = x + y (x + y)2 = x2 + 2xy + y2 (x + y)3 = x3 + 3x2 y + 3xy2 + y3 (x + y)4 = x4 + 4x3 y + 6x2 y2 + 4xy3 + y4 Can you guess the next expansion for the binomial (x + y)5? Figure 13.20 See Figure 13.20, which illustrates the following: • There are n + 1 terms in the expansion of (x + y) n. • The degree (or sum of the exponents) for
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each term is n. • The powers on x begin with n and decrease to 0. • The powers on y begin with 0 and increase to n. • The coefficients are symmetric. To determine the expansion on (x + y)5, we see n = 5, degree of 5. In descending order for powers of x, the pattern is as follows: thus, there will be 5+1 = 6 terms. Each term has a combined • • Introduce x5, and then for each successive term reduce the exponent on x by 1 until x0 = 1 is reached. Introduce y0 = 1, and then increase the exponent on y by 1 until y5 is reached. The next expansion would be (x + y)5 = x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5. x5, x4 y, x3 y2, x2 y3, xy4, y5 But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 13.21. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1517 Figure 13.21 To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add 1 + 1 to find the middle number, 2. In the nth row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it. To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form. The Binomial Theorem The Binomial Theorem is a formula that can be used to expand any binomial. (x + y) n n = ∑ k = 0 = xn + n ⎛ ⎞ ⎠xn − k yk ⎝ k n ⎛ ⎝ 1 ⎞ ⎠xn − 1 y + n ⎛ ⎝ 2 ⎞ ⎠xn − 2 y2 +... + n �
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�� ⎞ ⎠xyn − 1 + yn ⎝ n − 1 (13.12) Given a binomial, write it in expanded form. 1. Determine the value of n according to the exponent. 2. Evaluate the k = 0 through k = n using the Binomial Theorem formula. 3. Simplify. Example 13.38 Expanding a Binomial Write in expanded form. a. (x + y)5 1518 Chapter 13 Sequences, Probability, and Counting Theory b. ⎛ ⎝3x − y⎞ ⎠ 4 Solution a. Substitute n = 5 into the formula. Evaluate the k = 0 through k = 5 terms. Simplify. (x + y)5 = ⎞ ⎞ ⎠x2 y3 + ⎠x4 y1 + (x + y)5 = x5 + 5x4 y + 10x3 y2 + 10x2 y3 + 5xy4 + y5 ⎞ ⎠x5 y0 + ⎞ ⎠x3 y2 + ⎛ x1 y4 + ⎛ 5 ⎝ 4 ⎞ ⎠x0 y5 ⎛ 5 ⎝ 5 b. Substitute n = 4 into the formula. Evaluate the k = 0 through k = 4 terms. Notice that 3x is in the place that was occupied by x and that – y is in the place that was occupied by y. So we substitute them. Simplify. (3x − y)4 = ⎛ 4 ⎝ 2 (3x − y)4 = 81x4 − 108x3 y + 54x2 y2 − 12xy3 + y4 ⎞ ⎠(3x)3 ( − y)1 + ⎞ ⎠(3x)4 ( − y) ⎞ ⎠(3x)2 ( − y)2 + ⎞ ⎠(3x)1 ( − y)3 + ⎛ 4 ⎝ 3 ⎞ ⎠(3x)0 ( − y)4 ⎛ 4 ⎝ 4 Analysis Notice the alternating signs in part b. This happens because ( − y) raised to odd powers is negative, but ( − y) raised to even powers is positive.
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This will occur whenever the binomial contains a subtraction sign. 13.51 Write in expanded form. a. b. (x − y)5 (2x + 5y)3 Using the Binomial Theorem to Find a Single Term Expanding a binomial with a high exponent such as (x + 2y)16 can be a lengthy process. Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term. Note the pattern of coefficients in the expansion of (x + y)5. (x + y)5 = x5 + ⎞ ⎠x4 x3 y2 + ⎞ ⎠x2 y3 + ⎛ 5 ⎝ 3 ⎛ 5 ⎝ 4 ⎞ ⎠xy4 + y5 The second term is ⎛ 5 ⎝ 1 ⎠x4 y. The third term is ⎛ ⎞ 5 ⎝ 2 ⎞ ⎠x3 y2. We can generalize this result. n ⎛ r ⎝ ⎞ ⎠xn − r yr The (r+1)th Term of a Binomial Expansion The (r + 1)th term of the binomial expansion of (x + y) n is: n ⎛ r ⎝ ⎞ ⎠xn − r yr (13.13) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1519 Given a binomial, write a specific term without fully expanding. 1. Determine the value of n according to the exponent. 2. Determine (r + 1). 3. Determine r. 4. Replace r in the formula for the (r + 1)th term of the binomial expansion. Example 13.39 Writing a Given Term of a Binomial Expansion Find the tenth term of (x + 2y)16 without fully expanding the binomial. Solution Because we are looking for the tenth term, r + 1 = 10, we will use r = 9 in our calculations. n ⎛ r ⎝ ⎞ ⎠xn − r yr ⎛ 16 ⎝ 9 ⎞ ⎠x16 − 9 (2y
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)9 = 5,857,280x7 y9 13.52 Find the sixth term of (3x − y)9 without fully expanding the binomial. Access these online resources for additional instruction and practice with binomial expansion. • The Binomial Theorem (http://openstaxcollege.org/l/binomialtheorem) • Binomial Theorem Example (http://openstaxcollege.org/l/btexample) 1520 Chapter 13 Sequences, Probability, and Counting Theory 13.6 EXERCISES Verbal 322. What calculated? is a binomial coefficient, and how it is What role do binomial coefficients play in a binomial 323. expansion? Are they restricted to any type of number? 324. What is the Binomial Theorem and what is its use? When is it an advantage to use the Binomial 325. Theorem? Explain. Algebraic the following exercises, evaluate the binomial For coefficient. (3x − 2y)4 340. (4x − 3y)5 341. 5 ⎛ ⎝ x + 3y⎞ 1 ⎠ 342. (x−1 + 2y−1)4 343. ( x − y)5 For the following exercises, use the Binomial Theorem to write the first three terms of each binomial. 326. 327. 328. 329. 330. 331. 332 ⎛ 10 ⎝ 9 ⎞ ⎠ ⎛ 25 ⎝ 11 ⎞ ⎠ ⎛ 17 ⎝ 6 ⎞ ⎠ 333. ⎛ 200 ⎝ 199 ⎞ ⎠ For the following exercises, use the Binomial Theorem to expand each binomial. 334. (4a − b)3 335. (5a + 2)3 336. (3a + 2b)3 337. (2x + 3y)4 338. (4x + 2y)5 339. This content is available for free at https://cnx.org/content/col11758/1.5 344. (a + b)17 345. (x − 1)18 346. (a − 2b)15 347. (x − 2y)8 348. (3a + b)20 349. (2a + 4b)7 350. (x3 − y)8 For the following exercises
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, find the indicated term of each binomial without fully expanding the binomial. 351. 352. 353. 354. 355. The fourth term of (2x − 3y)4 The fourth term of (3x − 2y)5 The third term of (6x − 3y)7 The eighth term of (7 + 5y)14 The seventh term of (a + b)11 356. The fifth term of (x − y)7 The tenth term of (x − 1)12 The ninth term of (a − 3b2)11 357. 358. 359. Chapter 13 Sequences, Probability, and Counting Theory 1521 • • • ( a + 4 a − 5)8 (x3 + 2y2 − z)5 (3x2 − 2y3) 12 The fourth term of ⎛ ⎝x3 − 1 2 10 ⎞ ⎠ 360. The eighth term of ⎛ ⎝ 9 + 2 x ⎞ ⎠ y 2 Graphical For the following exercises, use the Binomial Theorem f (x) = (x + 3)4. Then find and to expand the binomial graph each indicated sum on one set of axes. 361. Find and graph f1(x), such that f1(x) is the first term of the expansion. 362. Find and graph f2(x), such that f2(x) is the sum of the first two terms of the expansion. 363. Find and graph f3(x), such that f3(x) is the sum of the first three terms of the expansion. 364. Find and graph f4(x), such that f4(x) is the sum of the first four terms of the expansion. 365. Find and graph f5(x), such that f5(x) is the sum of the first five terms of the expansion. Extensions n, each term has the 366. In the expansion of (5x + 3y) n form ⎛ ⎝ k ⎞ ⎠an – k bk n value 0, 1, 2,..., n. If ⎛ ⎝ k, where k successively takes on the ⎞ ⎠ = ⎞ ⎠, what ⎛ 7 ⎝ 2 is the corresponding term? 367. In the expansion of (a + b), the coefficient
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of an − k bk is the same as the coefficient of which other term? n 368. Consider the expansion of (x + b)40. What is the exponent of b in the kth term? 369. n ⎛ ⎝ k n ⎞ ⎞ Find ⎛ ⎠ and write ⎠ + ⎝ k − 1 n ⎞ binomial coefficient in the form ⎛ ⎠. Prove it. Hint: Use ⎝ k integer p, such that answer fact any the as a that, the for p ≥ 1, p! = p(p − 1)!. Which expression cannot be expanded using the 370. Binomial Theorem? Explain. • (x2 − 2x + 1) 1522 Chapter 13 Sequences, Probability, and Counting Theory 13.7 | Probability Learning Objectives In this section, you will: 13.7.1 Construct probability models. 13.7.2 Compute probabilities of equally likely outcomes. 13.7.3 Compute probabilities of the union of two events. 13.7.4 Use the complement rule to find probabilities. 13.7.5 Compute probability using counting theory. Figure 13.22 An example of a “spaghetti model,” which can be used to predict possible paths of a tropical storm.[1] Residents of the Southeastern United States are all too familiar with charts, known as spaghetti models, such as the one in Figure 13.22. They combine a collection of weather data to predict the most likely path of a hurricane. Each colored line represents one possible path. The group of squiggly lines can begin to resemble strands of spaghetti, hence the name. In this section, we will investigate methods for making these types of predictions. Constructing Probability Models Suppose we roll a six-sided number cube. Rolling a number cube is an example of an experiment, or an activity with an observable result. The numbers on the cube are possible results, or outcomes, of this experiment. The set of all possible outcomes of an experiment is called the sample space of the experiment. The sample space for this experiment is {1, 2, 3, 4, 5, 6}. An event is any subset of a sample space. The likelihood of an event is known as probability. The probability of an event p is a number that always satisfies 0 ≤ p ≤ 1, where 0 indicates an impossible event and 1 indicates a
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certain event. A probability model is a mathematical 1. The figure is for illustrative purposes only and does not model any particular storm. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1523 description of an experiment listing all possible outcomes and their associated probabilities. For instance, if there is a 1% chance of winning a raffle and a 99% chance of losing the raffle, a probability model would look much like Table 13.3. Outcome Probability Winning the raffle 1% Losing the raffle 99% Table 13.3 The sum of the probabilities listed in a probability model must equal 1, or 100%. Given a probability event where each event is equally likely, construct a probability model. 1. Identify every outcome. 2. Determine the total number of possible outcomes. 3. Compare each outcome to the total number of possible outcomes. Example 13.40 Constructing a Probability Model Construct a probability model for rolling a single, fair die, with the event being the number shown on the die. Solution Begin by making a list of all possible outcomes for the experiment. The possible outcomes are the numbers that can be rolled: 1, 2, 3, 4, 5, and 6. There are six possible outcomes that make up the sample space. Assign probabilities to each outcome in the sample space by determining a ratio of the outcome to the number of possible outcomes. There is one of each of the six numbers on the cube, and there is no reason to think that any particular face is more likely to show up than any other one, so the probability of rolling any number is 1 6. Outcome Roll of 1 Roll of 2 Roll of 3 Roll of 4 Roll of 5 Roll of 6 Probability Table 13.4 Do probabilities always have to be expressed as fractions? No. Probabilities can be expressed as fractions, decimals, or percents. Probability must always be a number between 0 and 1, inclusive of 0 and 1. 13.53 Construct a probability model for tossing a fair coin. 1524 Chapter 13 Sequences, Probability, and Counting Theory Computing Probabilities of Equally Likely Outcomes Let S be a sample space for an experiment. When investigating probability, an event is any subset of S. When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the
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event by the total number of outcomes in S. Suppose a number cube is rolled, and we are interested in finding the probability of the event “rolling a number less than or equal to 4.” There are 4 possible outcomes in the event and 6 possible outcomes in S, so the probability of the event is 4 6 = 2 3. Computing the Probability of an Event with Equally Likely Outcomes The probability of an event E in an experiment with sample space S with equally likely outcomes is given by P(E) = number of elements in E number of elements in S = n(E) n(S) (13.14) E is a subset of S, so it is always true that 0 ≤ P(E) ≤ 1. Example 13.41 Computing the Probability of an Event with Equally Likely Outcomes A number cube is rolled. Find the probability of rolling an odd number. Solution The event “rolling an odd number” contains three outcomes. There are 6 equally likely outcomes in the sample space. Divide to find the probability of the event. P(E) = 3 6 = 1 2 13.54 A number cube is rolled. Find the probability of rolling a number greater than 2. Computing the Probability of the Union of Two Events We are often interested in finding the probability that one of multiple events occurs. Suppose we are playing a card game, and we will win if the next card drawn is either a heart or a king. We would be interested in finding the probability of the next card being a heart or a king. The union of two events E and F, written E ∪ F, is the event that occurs if either or both events occur. Suppose the spinner in Figure 13.23 is spun. We want to find the probability of spinning orange or spinning a b. P(E ∪ F) = P(E) + P(F) − P(E ∩ F) This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1525 Figure 13.23 There are a total of 6 sections, and 3 of them are orange. So the probability of spinning orange is 3 6 of 6 sections, and 2 of them have a b. So the probability of spinning a b is 2 6 = 1 3 = 1 2. If we added these two probabilities,. There are a total we would be counting the sector that
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is both orange and a b twice. To find the probability of spinning an orange or a b, we need to subtract the probability that the sector is both orange and has a b The probability of spinning orange or a b is 2 3. Probability of the Union of Two Events The probability of the union of two events E and F (written E ∪ F ) equals the sum of the probability of E and the probability of F minus the probability of E and F occurring together ( which is called the intersection of E and F and is written as E ∩ F ). P(E ∪ F) = P(E) + P(F) − P(E ∩ F) (13.15) Example 13.42 Computing the Probability of the Union of Two Events A card is drawn from a standard deck. Find the probability of drawing a heart or a 7. Solution A standard deck contains an equal number of hearts, diamonds, clubs, and spades. So the probability of drawing a heart is 1. There are four 7s in a standard deck, and there are a total of 52 cards. So the probability of drawing 4 a 7 is 1 13. The only card in the deck that is both a heart and a 7 is the 7 of hearts, so the probability of drawing both a heart and a 7 is 1 52 and P(H ∩ 7) = 1 52. Substitute P(H) = 1 4 P(7) = 1 13 into the formula.,, 1526 Chapter 13 Sequences, Probability, and Counting Theory P(E ∪ F) = P(E) + P(F) − P(E ∩ F) − 1 52 + 1 13 = 1 4 = 4 13 The probability of drawing a heart or a 7 is 4 13. 13.55 A card is drawn from a standard deck. Find the probability of drawing a red card or an ace. Computing the Probability of Mutually Exclusive Events Suppose the spinner in Figure 13.23 is spun again, but this time we are interested in the probability of spinning an orange or a d. There are no sectors that are both orange and contain a d, so these two events have no outcomes in common. Events are said to be mutually exclusive events when they have no outcomes in common. Because there is no overlap, there is nothing to subtract, so the general formula is P(E ∪ F) = P(E) + P(F) Notice that
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with mutually exclusive events, the intersection of E and F is the empty set. The probability of spinning an orange is 3 6 and the probability of spinning a d is 1 6. We can find the probability of spinning an orange or a d simply = 1 2 by adding the two probabilities. P(E ∪ F) = P(E) + P(F The probability of spinning an orange or a d is 2 3. Probability of the Union of Mutually Exclusive Events The probability of the union of two mutually exclusive events E and F is given by P(E ∪ F) = P(E) + P(F) (13.16) Given a set of events, compute the probability of the union of mutually exclusive events. 1. Determine the total number of outcomes for the first event. 2. Find the probability of the first event. 3. Determine the total number of outcomes for the second event. 4. Find the probability of the second event. 5. Add the probabilities. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1527 Example 13.43 Computing the Probability of the Union of Mutually Exclusive Events A card is drawn from a standard deck. Find the probability of drawing a heart or a spade. Solution The events “drawing a heart” and “drawing a spade” are mutually exclusive because they cannot occur at the same time. The probability of drawing a heart is 1, so the 4, and the probability of drawing a spade is also 1 4 probability of drawing a heart or a spade is 1 4 + 1 4 = 1 2 13.56 A card is drawn from a standard deck. Find the probability of drawing an ace or a king. Using the Complement Rule to Compute Probabilities We have discussed how to calculate the probability that an event will happen. Sometimes, we are interested in finding the probability that an event will not happen. The complement of an event E, denoted E′, is the set of outcomes in the sample space that are not in E. For example, suppose we are interested in the probability that a horse will lose a race. If event W is the horse winning the race, then the complement of event W is the horse losing the race. To find the probability that the horse loses the race, we need to use the fact that the sum of all probabilities in
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a probability model must be 1. P(E′) = 1 − P(E) The probability of the horse winning added to the probability of the horse losing must be equal to 1. Therefore, if the probability of the horse winning the race is 1 9, the probability of the horse losing the race is simply 1 − 1 9 = 8 9 The Complement Rule The probability that the complement of an event will occur is given by P(E′) = 1 − P(E) (13.17) Example 13.44 Using the Complement Rule to Calculate Probabilities Two six-sided number cubes are rolled. a. Find the probability that the sum of the numbers rolled is less than or equal to 3. b. Find the probability that the sum of the numbers rolled is greater than 3. 1528 Chapter 13 Sequences, Probability, and Counting Theory Solution The first step is to identify the sample space, which consists of all the possible outcomes. There are two number cubes, and each number cube has six possible outcomes. Using the Multiplication Principle, we find that there are 6×6, or 36 total possible outcomes. So, for example, 1-1 represents a 1 rolled on each number cube. 1-1 1-2 1-3 1-4 1-5 1-6 2-1 2-2 2-3 2-4 2-5 2-6 3-1 3-2 3-3 3-4 3-5 3-6 4-1 4-2 4-3 4-4 4-5 4-6 5-1 5-2 5-3 5-4 5-5 5-6 6-1 6-2 6-3 6-4 6-5 6-6 Table 13.5 a. We need to count the number of ways to roll a sum of 3 or less. These would include the following outcomes: 1-1, 1-2, and 2-1. So there are only three ways to roll a sum of 3 or less. The probability is 3 36 b. Rather than listing all the possibilities, we can use the Complement Rule. Because we have already found the probability of the complement of this event, we can simply subtract that probability from 1 to find the probability that the sum of the numbers rolled is greater than 3. = 1 12 P(E′) = 1 − P(E) = 1 − 1 12 = 11 12 Two number cubes are rolled. Use the Complement
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Rule to find the probability that the sum is less than 13.57 10. Computing Probability Using Counting Theory Many interesting probability problems involve counting principles, permutations, and combinations. In these problems, we will use permutations and combinations to find the number of elements in events and sample spaces. These problems can be complicated, but they can be made easier by breaking them down into smaller counting problems. Assume, for example, that a store has 8 cellular phones and that 3 of those are defective. We might want to find the probability that a couple purchasing 2 phones receives 2 phones that are not defective. To solve this problem, we need to calculate all of the ways to select 2 phones that are not defective as well as all of the ways to select 2 phones. There are 5 phones that are not defective, so there are C(5, 2) ways to select 2 phones that are not defective. There are 8 phones, so there are C(8, 2) ways to select 2 phones. The probability of selecting 2 phones that are not defective is: This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1529 ways to select 2 phones that are not defective ways to select 2 phones = C(5, 2) C(8, 2) = 10 28 = 5 14 Example 13.45 Computing Probability Using Counting Theory A child randomly selects 5 toys from a bin containing 3 bunnies, 5 dogs, and 6 bears. a. Find the probability that only bears are chosen. b. Find the probability that 2 bears and 3 dogs are chosen. c. Find the probability that at least 2 dogs are chosen. Solution a. We need to count the number of ways to choose only bears and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 5) ways to choose 5 bears. There are 14 toys, so there are C(14, 5) ways to choose any 5 toys. C(6,5) C(14,5) = 6 2,002 = 3 1,001 b. We need to count the number of ways to choose 2 bears and 3 dogs and the total number of possible ways to select 5 toys. There are 6 bears, so there are C(6, 2) ways to choose 2 bears. There are 5 dogs, so there are C(5, 3) ways
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to choose 3 dogs. Since we are choosing both bears and dogs at the same time, we will use the Multiplication Principle. There are C(6, 2) ⋅ C(5, 3) ways to choose 2 bears and 3 dogs. We can use this result to find the probability. C(6,2)C(5,3) C(14,5) = 15 ⋅ 10 2,002 = 75 1,001 c. It is often easiest to solve “at least” problems using the Complement Rule. We will begin by finding the probability that fewer than 2 dogs are chosen. If less than 2 dogs are chosen, then either no dogs could be chosen, or 1 dog could be chosen. When no dogs are chosen, all 5 toys come from the 9 toys that are not dogs. There are C(9, 5) ways to choose toys from the 9 toys that are not dogs. Since there are 14 toys, there are C(14, 5) ways to choose the 5 toys from all of the toys. C(9,5) C(14,5) = 63 1,001 If there is 1 dog chosen, then 4 toys must come from the 9 toys that are not dogs, and 1 must come from the 5 dogs. Since we are choosing both dogs and other toys at the same time, we will use the Multiplication Principle. There are C(5, 1) ⋅ C(9, 4) ways to choose 1 dog and 1 other toy. C(5,1)C(9,4) C(14,5) = 5 ⋅ 126 2,002 = 315 1,001 Because these events would not occur together and are therefore mutually exclusive, we add the probabilities to find the probability that fewer than 2 dogs are chosen. We then subtract that probability from 1 to find the probability that at least 2 dogs are chosen. 63 1,001 + 315 1,001 = 378 1,001 1530 Chapter 13 Sequences, Probability, and Counting Theory 1 − 378 1,001 = 623 1,001 A child randomly selects 3 gumballs from a container holding 4 purple gumballs, 8 yellow gumballs, 13.58 and 2 green gumballs. a. Find the probability that all 3 gumballs selected are purple. b. Find the probability that no yellow gumballs are selected. c. Find the probability that at least 1 yellow gumball is selected. Access these
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online resources for additional instruction and practice with probability. • Introduction to Probability (http://openstaxcollege.org/l/introprob) • Determining Probability (http://openstaxcollege.org/l/determineprob) this website (http://openstaxcollege.org/l/PreCalcLPC11) Visit Learningpod. for additional practice questions from This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1531 13.7 EXERCISES Verbal 385. Find the probability of tossing two heads. What term is used to express the likelihood of an 371. event occurring? Are there restrictions on its values? If so, what are they? If not, explain. 386. Find the probability of tossing exactly one tail. 387. Find the probability of tossing at least one tail. 372. What is a sample space? 373. What is an experiment? For the following exercises, four coins are tossed. 388. What is the sample space? What is the difference between events and outcomes? 374. Give an example of both using the sample space of tossing a coin 50 times. 389. Find the probability of tossing exactly two heads. 390. Find the probability of tossing exactly three heads. The union of two sets is defined as a set of elements 375. that are present in at least one of the sets. How is this similar to the definition used for the union of two events from a probability model? How is it different? Numeric For the following exercises, use the spinner shown in Figure 13.24 to find the probabilities indicated. 391. Find the probability of tossing four heads or four tails. 392. Find the probability of tossing all tails. 393. Find the probability of tossing not all tails. Find the probability of tossing exactly two heads or at 394. least two tails. Find the probability of tossing either two heads or 395. three heads. For the following exercises, one card is drawn from a standard deck of 52 cards. Find the probability of drawing the following: Figure 13.24 376. Landing on red 377. Landing on a vowel 378. Not landing on blue 379. Landing on purple or a vowel 380. Landing on blue or a vowel 381. Landing on green or blue 382. Landing on yellow or a consonant 383. Not landing on yellow or a consonant For
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the following exercises, two coins are tossed. 384. What is the sample space? 396. A club 397. A two 398. Six or seven 399. Red six 400. An ace or a diamond 401. A non-ace 402. A heart or a non-jack For the following exercises, two dice are rolled, and the results are summed. Construct a table showing the sample space of 403. outcomes and sums. 404. Find the probability of rolling a sum of 3. Find the probability of rolling at least one four or a 405. sum of 8. Find the probability of rolling an odd sum less than 406. 9. Find the probability of rolling a sum greater than or 407. equal to 15. 1532 Chapter 13 Sequences, Probability, and Counting Theory 408. Find the probability of rolling a sum less than 15. Find the probability of rolling a sum less than 6 or 409. greater than 9. 425. How much less is a player’s chance of selecting 3 winning numbers than the chance of selecting either 4 or 5 winning numbers? Real-World Applications Find the probability of rolling a sum between 6 and 410. 9, inclusive. Use this data for the exercises that follow: In 2013, there were roughly 317 million citizens in the United States, and about 40 million were elderly (aged 65 and over).[2] 426. If you meet a U.S. citizen, what is the percent chance that the person is elderly? (Round to the nearest tenth of a percent.) 427. If you meet five U.S. citizens, what is the percent chance that exactly one is elderly? (Round to the nearest tenth of a percent.) 428. If you meet five U.S. citizens, what is the percent chance that three are elderly? (Round to the nearest tenth of a percent.) If you meet five U.S. citizens, what is the percent four are elderly? (Round to the nearest 429. chance that thousandth of a percent.) It is predicted that by 2030, one in five U.S. citizens 430. will be elderly. How much greater will the chances of meeting an elderly person be at that time? What policy changes do you foresee if these statistics hold true? 411. Find the probability of rolling a sum of 5 or 6. Find the probability of rolling any sum other than 5 412. or 6. For the following exercises, a coin is tossed, and a card is pulled from a standard deck. Find the probability of the following
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: 413. A head on the coin or a club 414. A tail on the coin or red ace 415. A head on the coin or a face card 416. No aces For the following exercises, use this scenario: a bag of M&Ms contains 12 blue, 6 brown, 10 orange, 8 yellow, 8 red, and 4 green M&Ms. Reaching into the bag, a person grabs 5 M&Ms. 417. What is the probability of getting all blue M&Ms? 418. What is the probability of getting 4 blue M&Ms? 419. What is the probability of getting 3 blue M&Ms? 420. What is the probability of getting no brown M&Ms? Extensions Use the following scenario for the exercises that follow: In the game of Keno, a player starts by selecting 20 numbers from the numbers 1 to 80. After the player makes his selections, 20 winning numbers are randomly selected from numbers 1 to 80. A win occurs if the player has correctly selected 3, 4, or 5 of the 20 winning numbers. (Round all answers to the nearest hundredth of a percent.) What 421. exactly 3 winning numbers? is the percent chance that a player selects What 422. exactly 4 winning numbers? is the percent chance that a player selects What is the percent chance that a player selects all 5 423. winning numbers? 424. What is the percent chance of winning? 2. United States Census Bureau. http://www.census.gov This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1533 CHAPTER 13 REVIEW KEY TERMS Addition Principle if one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways annuity an investment in which the purchaser makes a sequence of periodic, equal payments arithmetic sequence a sequence in which the difference between any two consecutive terms is a constant arithmetic series the sum of the terms in an arithmetic sequence binomial coefficient the number of ways to choose r objects from n objects where order does not matter; equivalent to n C(n, r), denoted ⎛ r ⎝ ⎞ ⎠ binomial expansion the result of expanding (x + y) n by multiplying Binomial Theorem a formula that can be used to expand any binomial combination a selection of objects in which order
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does not matter common difference the difference between any two consecutive terms in an arithmetic sequence common ratio the ratio between any two consecutive terms in a geometric sequence complement of an event the set of outcomes in the sample space that are not in the event E diverge a series is said to diverge if the sum is not a real number event any subset of a sample space experiment an activity with an observable result explicit formula a formula that defines each term of a sequence in terms of its position in the sequence finite sequence a function whose domain consists of a finite subset of the positive integers {1, 2, … n} for some positive integer n Fundamental Counting Principle if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways; also known as the Multiplication Principle geometric sequence a sequence in which the ratio of a term to a previous term is a constant geometric series the sum of the terms in a geometric sequence index of summation in summation notation, the variable used in the explicit formula for the terms of a series and written below the sigma with the lower limit of summation infinite sequence a function whose domain is the set of positive integers infinite series the sum of the terms in an infinite sequence lower limit of summation the number used in the explicit formula to find the first term in a series Multiplication Principle if one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways; also known as the Fundamental Counting Principle mutually exclusive events events that have no outcomes in common n factorial the product of all the positive integers from 1 to n nth partial sum the sum of the first n terms of a sequence nth term of a sequence a formula for the general term of a sequence outcomes the possible results of an experiment 1534 Chapter 13 Sequences, Probability, and Counting Theory permutation a selection of objects in which order matters probability a number from 0 to 1 indicating the likelihood of an event probability model a mathematical description of an experiment listing all possible outcomes and their associated probabilities recursive formula a formula that defines each term of a sequence using previous term(s) sample space the set of all possible outcomes of an experiment sequence a function whose domain is a subset of the positive integers series the sum of the terms in a sequence summation notation a notation for series using the Greek letter sigma; it includes an explicit formula and
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specifies the first and last terms in the series term a number in a sequence union of two events the event that occurs if either or both events occur upper limit of summation the number used in the explicit formula to find the last term in a series KEY EQUATIONS Formula for a factorial (n − 1)(n − 2) ⋯ (2)(1), for n ≥ 2 recursive formula for nth term of an arithmetic sequence an = an − 1 + dn ≥ 2 explicit formula for nth term of an arithmetic sequence an = a1 + d(n − 1) recursive formula for nth term of a geometric sequence an = ran − 1, n ≥ 2 explicit formula for nth term of a geometric sequence an = a1 r n − 1 sum of the first n terms of an arithmetic series sum of the first n terms of a geometric series sum of an infinite geometric series with – 1 < r < 1 Sn = n(a1 + an) 2 Sn = a1( Sn = a1 1 − r ⋅ r ≠ 1 This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1535 number of permutations of n distinct objects taken r at a time number of combinations of n distinct objects taken r at a time number of permutations of n non-distinct objects P(n, r) = n! (n − r)! C(n, r) = n! r!(n − r)! n! r1!r2! … rk! Binomial Theorem (x + yxn − k yk (r + 1)th term of a binomial expansion n ⎛ r ⎝ ⎞ ⎠xn − r yr probability of an event with equally likely outcomes P(E) = n(E) n(S) probability of the union of two events P(E ∪ F) = P(E) + P(F) − P(E ∩ F) probability of the union of mutually exclusive events P(E ∪ F) = P(E) + P(F) probability of the complement of an event P(E ') = 1 − P(E) KEY CONCEPTS 13.1 Sequences and Their Notations • A sequence is a list of numbers, called terms, written in a specific order. • Explicit
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formulas define each term of a sequence using the position of the term. See Example 13.1, Example 13.2, and Example 13.3. • An explicit formula for the nth term of a sequence can be written by analyzing the pattern of several terms. See Example 13.4. • Recursive formulas define each term of a sequence using previous terms. • Recursive formulas must state the initial term, or terms, of a sequence. • A set of terms can be written by using a recursive formula. See Example 13.5 and Example 13.6. • A factorial is a mathematical operation that can be defined recursively. • The factorial of n is the product of all integers from 1 to n See Example 13.7. 13.2 Arithmetic Sequences • An arithmetic sequence is a sequence where the difference between any two consecutive terms is a constant. • The constant between two consecutive terms is called the common difference. 1536 Chapter 13 Sequences, Probability, and Counting Theory • The common difference is the number added to any one term of an arithmetic sequence that generates the subsequent term. See Example 13.8. • The terms of an arithmetic sequence can be found by beginning with the initial term and adding the common difference repeatedly. See Example 13.9 and Example 13.10. • A recursive formula for an arithmetic sequence with common difference d is given by an = an − 1 + d, n ≥ 2. See Example 13.11. • As with any recursive formula, the initial term of the sequence must be given. • An explicit formula for an arithmetic sequence with common difference d is given by an = a1 + d(n − 1). See Example 13.12. • An explicit formula can be used to find the number of terms in a sequence. See Example 13.13. • In application problems, we sometimes alter the explicit formula slightly to an = a0 + dn. See Example 13.14. 13.3 Geometric Sequences • A geometric sequence is a sequence in which the ratio between any two consecutive terms is a constant. • The constant ratio between two consecutive terms is called the common ratio. • The common ratio can be found by dividing any term in the sequence by the previous term. See Example 13.15. • The terms of a geometric sequence can be found by beginning with the first term and multiplying by the common ratio repeatedly. See Example 13.16 and Example 13.18. • A recursive formula for a geometric sequence
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with common ratio r is given by an = ran – 1 for n ≥ 2. • As with any recursive formula, the initial term of the sequence must be given. See Example 13.17. • An explicit formula for a geometric sequence with common ratio r is given by an = a1 r n – 1. See Example 13.19. In application problems, we sometimes alter the explicit formula slightly to an = a0 r n •. See Example 13.20. 13.4 Series and Their Notations • The sum of the terms in a sequence is called a series. • A common notation for series is called summation notation, which uses the Greek letter sigma to represent the sum. See Example 13.21. • The sum of the terms in an arithmetic sequence is called an arithmetic series. • The sum of the first n terms of an arithmetic series can be found using a formula. See Example 13.22 and Example 13.23. • The sum of the terms in a geometric sequence is called a geometric series. • The sum of the first n terms of a geometric series can be found using a formula. See Example 13.24 and Example 13.25. • The sum of an infinite series exists if the series is geometric with –1 < r < 1. • If the sum of an infinite series exists, it can be found using a formula. See Example 13.26, Example 13.27, and Example 13.28. • An annuity is an account into which the investor makes a series of regularly scheduled payments. The value of an annuity can be found using geometric series. See Example 13.29. This content is available for free at https://cnx.org/content/col11758/1.5 Chapter 13 Sequences, Probability, and Counting Theory 1537 13.5 Counting Principles • • If one event can occur in m ways and a second event with no common outcomes can occur in n ways, then the first or second event can occur in m + n ways. See Example 13.30. If one event can occur in m ways and a second event can occur in n ways after the first event has occurred, then the two events can occur in m×n ways. See Example 13.31. • A permutation is an ordering of n objects. • If we have a set of n objects and we want to choose r objects from the set in order, we write P(n, r). • Permutation problems can be
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solved using the Multiplication Principle or the formula for P(n, r). See Example 13.32 and Example 13.33. • A selection of objects where the order does not matter is a combination. • Given n distinct objects, the number of ways to select r objects from the set is C(n, r) and can be found using a formula. See Example 13.34. • A set containing n distinct objects has 2 n subsets. See Example 13.35. • For counting problems involving non-distinct objects, we need to divide to avoid counting duplicate permutations. See Example 13.36. 13.6 Binomial Theorem • n ⎛ r ⎝ ⎞ ⎠ is called a binomial coefficient and is equal to C(n, r). See Example 13.37. • The Binomial Theorem allows us to expand binomials without multiplying. See Example 13.38. • We can find a given term of a binomial expansion without fully expanding the binomial. See Example 13.39. 13.7 Probability • Probability is always a number between 0 and 1, where 0 means an event is impossible and 1 means an event is certain. • The probabilities in a probability model must sum to 1. See Example 13.40. • When the outcomes of an experiment are all equally likely, we can find the probability of an event by dividing the number of outcomes in the event by the total number of outcomes in the sample space for the experiment. See Example 13.41. • To find the probability of the union of two events, we add the probabilities of the two events and subtract the probability that both events occur simultaneously. See Example 13.42. • To find the probability of the union of two mutually exclusive events, we add the probabilities of each of the events. See Example 13.43. • The probability of the complement of an event is the difference between 1 and the probability that the event occurs. See Example 13.44. • In some probability problems, we need to use permutations and combinations to find the number of elements in events and sample spaces. See Example 13.45. CHAPTER 13 REVIEW EXERCISES Sequences and Their Notation 431. Write the first four terms of the sequence defined by the recursive formula a1 = 2, an = an − 1 + n. 432. Evaluate 6! (5 − 3)!3!. 433. Write the first four terms of
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the sequence defined by n the explicit formula an = 10 + 3. 1538 Chapter 13 Sequences, Probability, and Counting Theory 434. Write the first four terms of the sequence defined by n! the explicit formula an = n(n + 1). Arithmetic Sequences 435. Is the sequence 4 7, 47 21 find the common difference., 82 21, 39 7,... arithmetic? If so, 448. Write a recursive formula for the geometric sequence 1, 1 3, 1 27, 1 9, … 449. Write an explicit formula for the geometric sequence − 1 5, − 1 135, − 1 15, − 1 45, … 436. Is the sequence 2, 4, 8, 16,... arithmetic? If so, find the common difference. −5 59,049? 450. How many terms are in the finite geometric sequence 437. An arithmetic sequence has the first term a1 = 18 and common difference d = − 8. What are the first five terms? Series and Their Notation 451. Use summation notation to write the sum of terms 1 2 m + 5 from m = 0 to m = 5. 438. An arithmetic sequence has terms a3 = 11.7 and a8 = − 14.6. What is the first term? 452. Use summation notation to write the sum that results from adding the number 13 twenty times. 439. Write a recursive formula for the arithmetic sequence −20, − 10, 0,10,… 453. Use the formula for the sum of the first n terms of an arithmetic series to find the sum of the first eleven terms of the arithmetic series 2.5, 4, 5.5, …. 440. Write a recursive formula for the arithmetic sequence, …, and then find the 31st term. 0, − 1 2, − 1, − 3 2 441. Write an explicit formula for the arithmetic sequence 7 8, 29 24, 37 24, 15 8, … 454. A ladder has 15 tapered rungs, the lengths of which increase by a common difference. The first rung is 5 inches long, and the last rung is 20 inches long. What is the sum of the lengths of the rungs? 455. Use the formula for the sum of the first n terms of a geometric series the series to find for S9 442. How many terms are in the finite arithmetic sequence 12, 20, 28, …, 172? 12
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, 6, 3, 3 2, … Geometric Sequences 443. Find the common ratio for the geometric sequence 2.5, 5, 10, 20, … 444. Is the sequence 4, 16, 28, 40, … geometric? If so find the common ratio. If not, explain why. 445. A geometric sequence has terms a7 = 16,384 and a9 = 262,144. What are the first five terms? 446. A geometric sequence has the first term a1 = − 3 and common ratio r = 1 2. What is the 8th term? 447. What are the first five terms of the geometric sequence a1 = 3, an = 4 ⋅ an − 1? This content is available for free at https://cnx.org/content/col11758/1.5 456. The fees for the first three years of a hunting club membership are given in Table 13.6. If fees continue to rise at the same rate, how much will the total cost be for the first ten years of membership? Year Membership Fees 1 2 3 $1500 $1950 $2535 Table 13.6 457. ∞ ∑ k = 1 Find the sum of the infinite geometric series 45 ⋅ ( − 1 3 ) k − 1. Chapter 13 Sequences, Probability, and Counting Theory 1539 458. A ball has a bounce-back ratio of 3 5 the height of the previous bounce. Write a series representing the total distance traveled by the ball, assuming it was initially dropped from a height of 5 feet. What is the total distance? (Hint: the total distance the ball travels on each bounce is the sum of the heights of the rise and the fall.) 459. Alejandro deposits $80 of his monthly earnings into an annuity that earns 6.25% annual interest, compounded monthly. How much money will he have saved after 5 years? 460. The twins Sarah and Scott both opened retirement accounts on their 21st birthday. Sarah deposits $4,800.00 each year, earning 5.5% annual interest, compounded monthly. Scott deposits $3,600.00 each year, earning 8.5% annual interest, compounded monthly. Which twin will earn the most interest by the time they are 55 years old? How much more? Counting Principles 461. How many ways are there to choose a number from the set { − 10, − 6,4,10,12,18,24
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,32} that is divisible by either 4 or 6? In a group of 20 musicians, 12 play piano, 7 play 462. trumpet, and 2 play both piano and trumpet. How many musicians play either piano or trumpet? 470. A day spa charges a basic day rate that includes use of a sauna, pool, and showers. For an extra charge, guests can choose from the following additional services: massage, body scrub, manicure, pedicure, facial, and straight-razor shave. How many ways are there to order additional services at the day spa? 471. How many distinct ways can the word DEADWOOD be arranged? 472. How many distinct rearrangements of the letters of the word DEADWOOD are there if the arrangement must begin and end with the letter D? Binomial Theorem 473. Evaluate the binomial coefficient ⎛ 23 ⎝ 8 ⎞ ⎠. 474. Use the Binomial Theorem to expand ⎛ ⎝3x + 1 2 6. y⎞ ⎠ 475. Use the Binomial Theorem to write the first three terms of (2a + b)17. 476. Find the fourth term of ⎛ ⎝3a2 − 2b⎞ ⎠ 11 without fully expanding the binomial. Probability 463. How many ways are there to construct a 4-digit code if numbers can be repeated? For the following exercises, assume two die are rolled. 477. Construct a table showing the sample space. 464. A palette of water color paints has 3 shades of green, 3 shades of blue, 2 shades of red, 2 shades of yellow, and 1 shade of black. How many ways are there to choose one shade of each color? 478. What is the probability that a roll includes a 2? 479. What is the probability of rolling a pair? 465. Calculate P(18, 4). 466. In a group of 5 freshman, 10 sophomores, 3 juniors, and 2 seniors, how many ways can a president, vice president, and treasurer be elected? 467. Calculate C(15, 6). 468. A coffee shop has 7 Guatemalan roasts, 4 Cuban roasts, and 10 Costa Rican roasts. How many ways can the shop choose 2 Guatemalan, 2 Cuban, and 3 Costa Rican roasts for a coffee tasting event
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? How 469. {1, 3, 5, …, 99} have? many subsets does the set 480. What is the probability that a roll includes a 2 or results in a pair? 481. What is the probability that a roll doesn’t include a 2 or result in a pair? 482. What is the probability of rolling a 5 or a 6? 483. What is the probability that a roll includes neither a 5 nor a 6? For the following exercises, use the following data: An elementary school survey found that 350 of the 500 students preferred soda to milk. Suppose 8 children from the (Show calculations and round to the nearest tenth of a percent.) attending a birthday party. school are 1540 Chapter 13 Sequences, Probability, and Counting Theory 484. What is the percent chance that all the children attending the party prefer soda? 485. What is the percent chance that at least one of the children attending the party prefers milk? 486. What is the percent chance that exactly 3 of the children attending the party prefer soda? 487. What is the percent chance that exactly 3 of the children attending the party prefer milk? CHAPTER 13 PRACTICE TEST 488. Write the first four terms of the sequence defined by the recursive formula a = – 14, an = 2 + an – 1 2. 498. Use summation notation to write the sum of terms 3k 2 − 5 6 k from k = − 3 to k = 15. 489. Write the first four terms of the sequence defined by the explicit formula an = n2 – n – 1 n!. 499. A community baseball stadium has 10 seats in the first row, 13 seats in the second row, 16 seats in the third row, and so on. There are 56 rows in all. What is the seating capacity of the stadium? 490. Is the sequence 0.3, 1.2, 2.1, 3, … arithmetic? If so find the common difference. 491. An arithmetic sequence has the first term a1 = − 4. What is the 6th term? and common difference d = – 4 3 492. Write a recursive formula for the arithmetic sequence, … and then find the 22nd −2, − 7 2, − 5, − 13 2 term. 493. Write an explicit formula for the arithmetic sequence 15.6, 15, 14.4, 13.8, … and then find the 32nd term
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. 494. Is the sequence − 2, − 1, − 1 2, − 1 4, … geometric? If so find the common ratio. If not, explain why. 495. What is the 11th term of the geometric sequence − 1.5, − 3, − 6, − 12, …? 496. Write a recursive formula for the geometric sequence 1, … 497. Write an explicit formula for the geometric sequence 4, − 4 3, − 4 27, 4 9, … This content is available for free at https://cnx.org/content/col11758/1.5 500. Use the formula for the sum of the first n terms of a 7 geometric series to find ∑ k = 1 −0.2 ⋅ (−5) k − 1. 501. ∞ ∑ k = 1 1 3 Find the sum of k − 1 ⋅ ⎛ ⎝− 1 5 ⎞ ⎠. the infinite geometric series 502. Rachael deposits $3,600 into a retirement fund each year. The fund earns 7.5% annual interest, compounded monthly. If she opened her account when she was 20 years old, how much will she have by the time she’s 55? How much of that amount was interest earned? 503. In a competition of 50 professional ballroom dancers, 22 compete in the fox-trot competition, 18 compete in the tango competition, and 6 compete in both the fox-trot and tango competitions. How many dancers compete in the foxtrot or tango competitions? 504. A buyer of a new sedan can custom order the car by choosing from 5 different exterior colors, 3 different interior colors, 2 sound systems, 3 motor designs, and either manual or automatic transmission. How many choices does the buyer have? 505. To allocate annual bonuses, a manager must choose his top four employees and rank them first to fourth. In how many ways can he create the “Top-Four” list out of the 32 employees? Chapter 13 Sequences, Probability, and Counting Theory 1541 506. A rock group needs to choose 3 songs to play at the annual Battle of the Bands. How many ways can they choose their set if have 15 songs to pick from? chance that exactly 3 are butterscotch? (Show calculations and round to the nearest tenth of a percent.) 507. A self-serve frozen yogurt shop
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has 8 candy toppings and 4 fruit toppings to choose from. How many ways are there to top a frozen yogurt? How many 508. the word EVANESCENCE be arranged if the anagram must end with the letter E? distinct ways can 509. Use the Binomial Theorem to expand ⎛ ⎝ 3 2 x − 1 2 5. y⎞ ⎠ 510. Find the seventh term of ⎛ ⎝x2 − 1 2 expanding the binomial. 13 ⎞ ⎠ without fully For the following exercises, use the spinner in Figure 13.25. Figure 13.25 511. Construct a probability model showing each possible outcome and its associated probability. (Use the first letter for colors.) 512. What number? is the probability of landing on an odd 513. What is the probability of landing on blue? 514. What is the probability of landing on blue or an odd number? 515. What is the probability of landing on anything other than blue or an odd number? 516. A bowl of candy holds 16 peppermint, 14 butterscotch, and 10 strawberry flavored candies. Suppose a person grabs a handful of 7 candies. What is the percent 1542 Chapter 13 Sequences, Probability, and Counting Theory This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1543 A | APPENDIX A1 Appendix Important Proofs and Derivations Product Rule loga xy = loga x + loga y Proof: Let m = loga x and n = loga y. Write in exponent form. x = am and y = an. Multiply. xy = am an = am + n am + n = xy loga(xy) = m + n = loga x + logb y Change of Base Rule loga b = logc b logc a loga b = 1 logb a where x and y are positive, and a > 0, a ≠ 1. Proof: Let x = loga b. Write in exponent form. a x = b Take the logc of both sides. logc a x = logc b xlogc a = logc b logc b logc a logc b loga b loga b = x = When c = b, loga b = logb b logb a = 1
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logb a Heron’s Formula A = s(s − a)(s − b)(s − c) 1544 where s = a + b + c 2 Proof: Let a, b, and c be the sides of a triangle, and h be the height. Appendix A Figure A1 So s = a + b + c 2. We can further name the parts of the base in each triangle established by the height such that p + q = c. Figure A2 Using the Pythagorean Theorem, h2 + p2 = a2 and h2 + q2 = b2. Since q = c − p, then q2 = (c − p)2. Expanding, we find that q2 = c2 − 2cp + p2. We can then add h2 to each side of the equation to get h2 + q2 = h2 + c2 − 2cp + p2. Substitute this result into the equation h2 + q2 = b2 yields b2 = h2 + c2 − 2cp + p2. Then replacing h2 + p2 with a2 gives b2 = a2 − 2cp + c2. Solve for p to get p = a2 + b2 − c2 2c Since h2 = a2 − p2, we get an expression in terms of a, b, and c. This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A h2 = a2 − p2 1545 = (a + p)(a − p) ⎝a2 + c2 − b2⎞ ⎛ ⎠ 2c = ⎝a2 + c2 − b2⎞ ⎛ ⎠ 2c ⎡ ⎤ ⎤ ⎡ ⎢a + ⎥ ⎥ ⎢a − ⎣ ⎦ ⎦ ⎣ ⎝2ac − a2 − c2 + b2⎞ ⎛ ⎝2ac + a2 + c2 − b2⎞ ⎛ ⎠ ⎠ 4c2 ⎝b2 − (a − c)2⎞ ⎛ ⎝(a + c)2 − b2⎞ ⎛ ⎠ ⎠
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4c2 (a + b + c)(a + c − b)(b + a − c)(b − a + c) 4c2 (a + b + c)( − a + b + c)(a − b + c)(a + b − c) 4c2 2s ⋅ (2s − a) ⋅ (2s − b)(2s − c) 4c2 = = = = = Therefore, h2 = h = 4s(s − a)(s − b)(s − c) c2 2 s(s − a)(s − b)(s − c) c And since A = 1 2 ch, then A = 1 2 c2 s(s − a)(s − b)(s − c) c = s(s − a)(s − b)(s − c) Properties of the Dot Product u · v = v · u Proof: u · v = 〈 u1, u2,...un 〉 · 〈 v1, v2,...vn 〉 = u1 v1 + u2 v2 +... + un vn = v1 u1 + v2 u2 +... + vn vn = 〈 v1, v2,...vn 〉 · 〈 u1, u2,...un 〉 = v · u u · (v + w) = u · v + u · w Proof: u · (v + w) = 〈 u1, u2,...un 〉 · ⎛ ⎝ 〈 v1, v2,...vn 〉 + 〈 w1, w2,...wn 〉 ⎞ ⎠ = 〈 u1, u2,...un 〉 · 〈 v1 + w1, v2 + w2,...vn + wn 〉 = 〈 u1(v1 + w1), u2(v2 + w2),...un(vn + wn) 〉 = 〈 u1 v1 + u1 w1, u2 v2 + u2 w2,...un vn + un wn 〉 = 〈 u1 v1, u2 v2,..., un vn 〉 + 〈
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u1 w1, u2 w2,..., un wn 〉 = 〈 u1, u2,...un 〉 · 〈 v1, v2,...vn 〉 + 〈 u1, u2,...un 〉 · 〈 w1, w2,...wn 〉 = u|2 Proof: 1546 Appendix A u · u = 〈 u1, u2,...un 〉 · 〈 u1, u2,...un 〉 = u1 u1 + u2 u2 +... + un un 2 + u2 2 +... + un = u1 = | 〈 u1, u2,...un 〉 |2 = v · u 2 Standard Form of the Ellipse centered at the Origin y2 b2 1 = x2 a2 + Derivation An ellipse consists of all the points for which the sum of distances from two foci is constant: (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 + (x − c)2 + ⎛ ⎝y − 0⎞ ⎠ 2 = constant Figure A3 Consider a vertex. Figure A4 Then, (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 + (x − c)2 + ⎛ ⎝y − 0⎞ ⎠ 2 = 2a Consider a covertex. This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1547 Figure A5 Then b2 + c2 = a2. (x − ( − c))2 + (y − 0)2 + (x − c)2 + (y − 0)2 = 2a (x + c)2 + y2 = 2a − (x − c)2 + y2 ⎝2a − (x − c)2 + y2⎞ ⎛ ⎠ (x + c)2 + y2 = 2 x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + (x − c)2 + y2 x2 + 2cx + c
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2 + y2 = 4a2 − 4a (x − c)2 + y2 + x2 − 2cx + y2 2cx = 4a2 − 4a (x − c)2 + y2 − 2cx − 1 4a ⎠ = (x − c)2 + y2 4cx − 4a2 = 4a (x − c)2 + y2 ⎛ ⎝4cx − 4a2⎞ c ax = (x − c)2 + y2 x2 = (x − c)2 + y2 a2 − 2xc + a2 − 2xc + a − c2 a2 c2 a2 c2 a2 c2 a2 a2 + a2 + x2 = x2 − 2xc + c2 + y2 x2 = x2 + c2 + y2 x2 = x2 + c2 + y2 Let 1 = a2 a2. a2 − c2 = x2 − c2 a2 x2 + y2 a2 − c2 = x2 ⎛ ⎝1 − c2 a2 ⎞ ⎠ + y2 a2 − c2 = x2 ⎛ a2 − c2 ⎝ a2 ⎞ ⎠ + y2 1 = x2 a2 + y2 a2 − c2 (A1) 1548 Because b2 + c2 = a2, then b2 = a2 − c2. Standard Form of the Hyperbola 1 = 1 = x2 a2 + x2 a2 + y2 a2 − c2 y2 b2 Appendix A (A2) y2 b2 1 = x2 a2 − Derivation A hyperbola is the set of all points in a plane such that the absolute value of the difference of the distances between two fixed points is constant. Figure A6 Diagram 1: The difference of the distances from Point P to the foci is constant: (x − ( − c))2 + (y − 0)2 − (x − c)2 + (y − 0)2 = constant Diagram 2: When the point is a vertex, the difference is 2a. (x − (−c))2 + ⎛ ⎝y − 0⎞ ⎠ 2 − (x − c)2 + �
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�� ⎝y − 0⎞ ⎠ 2 = 2a This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1549 (x − ( − c))2 + (y − 0)2 − (x − c)2 + (y − 0)2 = 2a (x + c)2 + y2 − (x − c)2 + y2 = 2a (x + c)2 + y2 = 2a + (x − c)2 + y2 ⎝2a + (x − c)2 + y2⎞ ⎛ (x + c)2 + y2 = ⎠ x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 x2 + 2cx + c2 + y2 = 4a2 + 4a (x − c)2 + y2 + x2 − 2cx + y2 2cx = 4a2 + 4a (x − c)2 + y2 − 2cx 4cx − 4a2 = 4a (x − c)2 + y2 cx − a2 = a (x − c)2 + y2 2 ⎝cx − a2⎞ ⎛ ⎠ = a2 ⎛ ⎝(x − c)2 + y2⎞ ⎠ c2 x2 − 2a2 c2 x2 + a4 = a2 x2 − 2a2 c2 x2 + a2 c2 + a2 y2 c2 x2 + a4 = a2 x2 + a2 c2 + a2 y2 a4 − a2 c2 = a2 x2 − c2 x2 + a2 y2 a2 ⎛ ⎝a2 − c2⎞ ⎝a2 − c2⎞ ⎠x2 + a2 y2 ⎠x2 − a2 y2 ⎛ ⎝a2 − c2⎞ ⎛ ⎝c2 − a2⎞ ⎠ = ⎠ = a2 ⎛ Define b as a positive number such that b2 = c2 − a2. (A3
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) a2 b2 = b2 x2 − a2 y2 a2 y2 a2 b2 a2 b2 = a2 b2 b2 x2 a2 b2 − y2 x2 a2 − b2 Trigonometric Identities 1 = Pythagorean Identity Even-Odd Identities Table A1 cos2 t + sin2 t = 1 1 + tan2 t = sec2 t 1 + cot2 t = csc2 t cos( − t) = cos t sec( − t) = sec t sin( − t) = − sin t tan( − t) = − tan t csc( − t) = − csc t cot( − t) = − cot t 1550 Appendix A ⎛ ⎝ tan t = cot ⎛ cos t = sin ⎝ π 2 π ⎛ sin t = cos ⎝ 2 π 2 π 2 π ⎛ sec t = csc ⎝ 2 π 2 ⎛ csc t = sec ⎝ ⎛ cot t = tan ⎝ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ − t⎞ ⎠ tan t = sin t cos t sec t = 1 cos t csc t = 1 sin t tan t = cos t cot t = 1 sin t cos(α + β) = cos α cos β − sin α sin β cos(α − β) = cos α cos β + sin α sin β sin(α + β) = sin α cos β + cos α sin β sin(α − β) = sin α cos β − cos α sin β tan(α + β) = tan(α − β) = tan α + tan β 1 − tan α tan β tan α − tan β 1 + tan α tan β sin(2θ) = 2sin θ cos θ cos(2θ) = cos2 θ − sin2 θ cos(2θ) = 1 − 2sin2 θ cos(2θ) = 2cos2 θ − 1 tan(2θ) = 2tan θ 1 − tan2 θ sin cos tan tan tan − cos α = ± 1 + cos α =
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± 1 − cos α 1 + cos α = sin α 1 + cos α = 1 − cos α sin α Cofunction Identities Fundamental Identities Sum and Difference Identities Double-Angle Formulas Half-Angle Formulas Table A1 This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1551 Reduction Formulas sin2 θ = cos2 θ = tan2 θ = 1 − cos(2θ) 2 1 + cos(2θ) 2 1 − cos(2θ) 1 + cos(2θ) Product-to-Sum Formulas Sum-to-Product Formulas ⎡ ⎡ cosαcosβ = 1 2 sinαcosβ = 1 ⎣sin⎛ 2 sinαsinβ = 1 ⎣cos⎛ 2 cosαsinβ = 1 2 ⎣sin⎛ ⎡ ⎡ ⎣cos⎛ ⎝α − β⎞ ⎠ + cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ ⎝α + β⎞ ⎠ + sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ ⎝α − β⎞ ⎠ − cos⎛ ⎝α + β⎞ ⎤ ⎦ ⎠ ⎝α + β⎞ ⎠ − sin⎛ ⎝α − β⎞ ⎤ ⎦ ⎠ ⎛ sinα + sinβ = 2sin ⎝ ⎛ sinα − sinβ = 2sin ⎝ cos ⎝ ⎠ cos ⎝ ⎠ 2 2 α − β α + β ⎛ ⎞ ⎛ cosα − cosβ = − 2sin ⎠sin ⎝ ⎝ cos ⎝ 2 2 ⎛ cosα + cosβ = 2cos ⎝ ⎞ ⎠ ⎞ ⎠ Law of Sines Law of Cosines Table A1 sin α a = a sin α = sin β b = b sin β = sin γ c c sin γ a2 = b2 + c2 − 2bc cos α b
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2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ 1552 Appendix A ToolKit Functions Figure A7 Figure A8 This content is available for free at https://cnx.org/content/col11758/1.5 Appendix A 1553 Figure A9 Trigonometric Functions Unit Circle Figure A10 1554 Appendix A Angle 0 π 6, or 30 ° π 4, or 45 ° π 3, or 60 ° π 2, or 90 ° Cosine Sine Tangent Secant 1 0 0 1 Cosecant Undefined Cotangent Undefined Table A2 Undefined Undefined 1 0 This content is available for free at https://cnx.org/content/col11758/1.5 Index 1555 INDEX A absolute maximum, 294, 379 absolute minimum, 294, 379 absolute value, 272, 353 absolute value equation, 197, 216 absolute value functions, 353, 357 absolute value inequality, 210 addition method, 1217, 1224, 1331 Addition Principle, 1502, 1533 addition property, 206 adjacent side, 835, 892 algebraic expression, 23, 99 altitude, 1060, 1201 ambiguous case, 1063, 1201 amplitude, 907, 967 angle, 810, 892 angle of depression, 843, 892 angle of elevation, 843, 892, 1060 angle of rotation, 1409, 1437 angular speed, 828, 892 annual interest, 1497 annual percentage rate (APR), 656, 793 annuity, 1497, 1533 apoapsis, 1422 arc, 815 arc length, 816, 825, 851, 892 arccosine, 951, 967 Archimedes’ spiral, 1129, 1201 arcsine, 951, 967 arctangent, 951, 967 area, 152, 216 area of a circle, 499 area of a sector, 827, 892 argument, 1138, 1201 arithmetic sequence, 1463, 1465, 1467, 1467, 1488, 1533 arithmetic series, 1489, 1533 arrow notation, 580, 634 associative property of addition, 20, 99 associative property of multiplication, 19, 99 asymptotes, 1365 augmented matrix, 1286, 1290, 1292, 1306, 1331
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average rate of change, 282, 379 axes of symmetry, 1365 axis of symmetry, 476, 481, 634, 1395, 1397 B base, 16, 99 binomial, 68, 99, 552 binomial coefficient, 1514, 1533 binomial expansion, 1515, 1518, 1533 Binomial Theorem, 1517, 1533 break-even point, 1225, 1331 C cardioid, 1118, 1201 carrying capacity, 762, 793 Cartesian coordinate system, 108, 216 Cartesian equation, 1103 Celsius, 362 center of a hyperbola, 1365, 1437 center of an ellipse, 1344, 1437 central rectangle, 1365 change-of-base formula, 732, 793 circle, 1249, 1250 circular motion, 917 circumference, 815 co-vertex, 1344 co-vertices, 1345 coefficient, 68, 99, 500, 564, 634 coefficient matrix, 1287, 1289, 1308, 1331 cofunction, 1000 cofunction identities, 841, 1000 column, 1272, 1331 column matrix, 1273 combination, 1533 combinations, 1508, 1514 combining functions, 301 common base, 738 common difference, 1463, 1488, 1533 common logarithm, 690, 793 common ratio, 1476, 1491, 1533 commutative, 303 commutative property of addition, 19, 99 commutative property of multiplication, 19, 99 complement of an event, 1527, 1533 completing the square, 179, 181, 216 complex conjugate, 166, 216 Complex Conjugate Theorem, 570 complex number, 161, 216, 1134 complex plane, 161, 216, 1134 composite function, 302, 379 composition of functions, 301 compound inequality, 209, 216 compound interest, 656, 793 compression, 410, 674, 707 conditional equation, 129, 216 conic, 1342, 1364, 1431 conic section, 1437 conic sections, 1159 conjugate axis, 1365, 1437 consistent system, 1213, 1331 constant, 23, 99 constant of variation, 625, 634 continuous, 525 continuous function, 516, 634 convex limaçons, 1120 convex limaҫon, 1201 coordinate plane, 1387 correlation coefficient, 457, 4
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66 cosecant, 874, 892, 932 cosecant function, 932, 933, 979 cosine, 1022, 1024 cosine function, 851, 892, 903, 905, 907, 917 cost function, 300, 1225, 1331 cotangent, 873, 892, 940 cotangent function, 940, 979 coterminal angles, 821, 824, 892 Cramer’s Rule, 1318, 1321, 1326, 1331 cube root, 500 cubic functions, 612 curvilinear path, 1148 D de Moivre, 1141 De Moivre’s Theorem, 1143, 1144, 1201 decompose a composite function, 311 decomposition, 1260 decreasing function, 289, 379, 397 decreasing linear function, 397, 466 1556 Index degenerate conic sections, 1404, 1437 degree, 68, 99, 507, 634, 811, 892 dependent system, 1213, 1224, 1239, 1331 dependent variable, 226, 379 Descartes, 1134 Descartes’ Rule of Signs, 572, 634 determinant, 1317, 1320, 1321, 1331 difference of squares, 73, 99 dimpled limaçons, 1120 dimpled limaҫon, 1201 direct variation, 625, 634 directrix, 1387, 1393, 1395, 1422, 1430, 1431, 1437 discriminant, 183, 216 displacement, 828 distance formula, 118, 216, 991, 1366, 1388 distributive property, 20, 99 diverge, 1533 diverges, 1493 dividend, 551 Division Algorithm, 552, 562, 634 divisor, 551 domain, 227, 237, 256, 258, 379, 950 domain and range, 256 domain of a composite function, 309 dot product, 1193, 1201 double-angle formulas, 1007, 1049 doubling time, 759, 793 Dürer, 1122 E eccentricity, 1423, 1437 electrostatic force, 286 elimination, 1250 ellipse, 1168, 1250, 1343, 1344, 1345, 1348, 1354, 1386, 1424, 1430, 1437
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ellipsis, 1446 end behavior, 502, 592, 634 endpoint, 285, 810 entry, 1273, 1331 equation, 26, 99, 235 equation in quadratic form, 199 equation in two variables, 111, 216 equations in quadratic form, 216 Euler, 1134 even function, 334, 379, 978 even-odd identities, 977, 1049 event, 1522, 1533 experiment, 1522, 1533 explicit formula, 1446, 1468, 1480, 1533 exponent, 16, 99 Exponential decay, 645 exponential decay, 653, 667, 753, 756, 760, 774 exponential equation, 737 exponential function, 645 exponential growth, 645, 649, 668, 753, 759, 762, 793 exponential notation, 16, 99 extraneous solution, 743, 793 extraneous solutions, 194, 216 extrapolation, 454, 466 F factor by grouping, 81, 99 Factor Theorem, 563, 634 factorial, 1457 factoring, 172 Fahrenheit, 362 feasible region, 1254, 1331 finite arithmetic sequence, 1469 finite sequence, 1448, 1533 foci, 1343, 1344, 1345, 1366, 1437 focus, 1343, 1387, 1393, 1395, 1422, 1430, 1431 focus (of a parabola), 1437 focus (of an ellipse), 1437 formula, 27, 99, 235 function, 227, 273, 379 function notation, 229 Fundamental Counting Principle, 1504, 1533 Fundamental Theorem of Algebra, 568, 570, 634 G Gauss, 1134, 1233, 1286 Gaussian elimination, 1233, 1289, 1331 general form, 478 general form of a quadratic function, 481, 634 Generalized Pythagorean Theorem, 1079, 1201 geometric sequence, 1476, 1491, 1533 geometric series, 1492, 1533 global maximum, 541, 634 This content is available for free at https://cnx.org/content/col11758/1.5 global minimum, 541, 634 graph in two variables, 111, 216 gravity, 1172 greatest common factor, 78, 99, 173 H half-angle formulas, 1014, 1049 half-life, 748, 753
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, 793 Heaviside method, 1262 Heron of Alexandria, 1086 Heron’s formula, 1086 horizontal asymptote, 583, 590, 634 horizontal compression, 340, 379, 1042 horizontal line, 140, 418, 466 horizontal line test, 244, 379 horizontal reflection, 328, 379 horizontal shift, 321, 379, 671, 702, 904 horizontal stretch, 340, 379 hyperbola, 1364, 1369, 1370, 1371, 1375, 1377, 1380, 1387, 1425, 1428, 1437 hypotenuse, 835, 892 I identities, 882, 892 identity equation, 129, 216 identity matrix, 1301, 1306, 1331 identity property of addition, 21, 99 identity property of multiplication, 21, 99 imaginary number, 161, 216 inconsistent equation, 130, 216 inconsistent system, 1213, 1222, 1238, 1331 increasing function, 289, 379, 397 increasing linear function, 397, 466 independent system, 1213, 1331 independent variable, 226, 379 index, 61, 99 index of summation, 1488, 1533 inequality, 1252 infinite geometric series, 1493 infinite sequence, 1448, 1533 infinite series, 1493, 1533 initial point, 1178, 1182, 1201 initial side, 811, 892 inner-loop limaçon, 1201 inner-loop limaçons, 1122 Index 1558 input, 227, 379 integers, 10, 15, 99 intercepts, 116, 216 Intermediate Value Theorem, 537, 634 interpolation, 454, 466 intersection, 1525 interval, 204, 216 interval notation, 204, 216, 257, 290, 379 inverse cosine function, 951, 967 inverse function, 364, 379, 613 inverse matrix, 1306, 1308 inverse of a radical function, 616 inverse of a rational function, 619 inverse property of addition, 21, 99 inverse property of multiplication, 21, 100 inverse sine function, 951, 967 inverse tangent function, 951, 967 inverse trigonometric functions, 950, 951, 955, 960 inverse variation, 627, 634 inverse variations, 627 inversely proportional, 627, 634 invertible function, 634 invertible functions, 610 invertible matrix, 1301
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, 1317 irrational numbers, 12, 15, 100 J joint variation, 629, 634 K Kronecker, 1134 L latus rectum, 1387, 1395, 1437 Law of Cosines, 1080, 1201 Law of Sines, 1062, 1081, 1201 leading coefficient, 68, 100, 507, 634 leading term, 68, 100, 507, 635 least common denominator, 92, 100, 131 least squares regression, 455, 466 lemniscate, 1124, 1201 linear equation, 129, 216 Linear Factorization Theorem, 570, 635 linear function, 394, 396, 466 linear growth, 645 linear inequality, 216 linear model, 438, 451 linear relationship, 451 linear speed, 829, 892 local extrema, 288, 379 local maximum, 288, 379, 541 local minimum, 288, 379, 541 logarithm, 687, 793 logarithmic equation, 744 logarithmic model, 778 logistic growth model, 762, 793 long division, 550 lower limit of summation, 1488, 1533 M magnitude, 272, 319, 1135, 1178, 1201 main diagonal, 1288, 1331 major and minor axes, 1345 major axis, 1344, 1348, 1437 matrix, 1272, 1273, 1286, 1331 matrix multiplication, 1279, 1303, 1308 matrix operations, 1273 maximum value, 476 measure of an angle, 811, 892 midline, 907, 967 midpoint formula, 122, 216 minimum value, 476 minor axis, 1344, 1437 model breakdown, 453, 466 modulus, 272, 1138, 1201 monomial, 68, 100 Multiplication Principle, 1504, 1533 multiplication property, 206 multiplicative inverse, 1304, 1304 multiplicative inverse of a matrix, 1301, 1331 multiplicity, 530, 635 mutually exclusive events, 1526, 1533 N n factorial, 1457, 1533 natural logarithm, 692, 742, 793 natural numbers, 10, 15, 100, 226 negative angle, 812, 892 Newton’s Law of Cooling, 760, 793 nominal rate, 656, 793 non-right triangles, 1060 nondegenerate conic section
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, 1406, 1437 nondegenerate conic sections, 1404 nonlinear inequality, 1252, 1331 nth term of the sequence, 1446 nth partial sum, 1533 nth root of a complex number, 1144 nth term of a sequence, 1533 nth partial sum, 1487 O oblique triangle, 1060, 1201 odd function, 334, 379, 977 one-loop limaçon, 1120 one-loop limaҫon, 1201 one-to-one, 668, 686, 723, 732 one-to-one function, 241, 364, 379, 950 opposite side, 835, 892 order of magnitude, 754, 793 order of operations, 16, 100 ordered pair, 109, 216, 226, 258 ordered triple, 1233 origin, 109, 216, 354 outcomes, 1522, 1533 output, 227, 380 P parabola, 476, 486, 1165, 1246, 1386, 1393, 1397, 1422, 1426, 1437 parallel, 141 parallel lines, 421, 466 parallelograms, 1184 parameter, 1148, 1201 parametric equations, 1148, 1164, 1164 parametric form, 1168 parent function, 702 partial fraction decomposition, 1260, 1331 partial fractions, 1260, 1331 Pascal, 1122 Pascal's Triangle, 1516 perfect square trinomial, 72, 100 periapsis, 1422 perimeter, 152, 217 period, 885, 892, 904, 925, 927, 1031 periodic function, 904, 967 permutation, 1504, 1534 perpendicular, 141 perpendicular lines, 421, 466 pH, 722 1559 Index phase shift, 908, 967 piecewise function, 273, 380 piecewise functions, 1451 point-slope form, 401, 466 point-slope formula, 143, 1372 polar axis, 1095, 1201 polar coordinates, 1095, 1097, 1100, 1111, 1202 polar equation, 1104, 1112, 1114, 1202, 1423, 1437 polar form, 1135 polar form of a complex number, 1137, 1202 polar form of a conic, 1432 polar grid, 1095 pole, 1095, 1202 polynomial, 68, 100, 564 polyn
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omial equation, 192, 217 polynomial function, 506, 523, 533, 539, 635 position vector, 1179, 1181 positive angle, 812, 892 power function, 500, 635 power rule for logarithms, 726, 732, 793 principal nth root, 61, 100 principal square root, 54, 100 probability, 1522, 1534 probability model, 1522, 1534 product of two matrices, 1279 product rule for logarithms, 724, 726, 793 product-to-sum formula, 1049 product-to-sum formulas, 1022, 1024 profit function, 1226, 1331 properties of determinants, 1324 Proxima Centauri, 754 Pythagoras, 1134 Pythagorean identities, 976, 1049 Pythagorean Identity, 854, 881, 892 Pythagorean identity, 991 Pythagorean Theorem, 118, 184, 217, 1009, 1043, 1079, 1170 Q quadrant, 108, 217 quadrantal angle, 813, 892 quadratic, 1264, 1266 quadratic equation, 172, 179, 181, 217, 1037 quadratic formula, 181, 183, 217, 1037 quadratic function, 481, 485 quotient, 551 quotient identities, 979, 1049 quotient rule for logarithms, 725, 793 R radian, 815, 817, 817, 892 radian measure, 817, 892 radical, 54, 100 radical equation, 194, 217 radical expression, 54, 100 radical functions, 610 radicand, 54, 100, 194 radiocarbon dating, 757 range, 227, 380, 952 rate of change, 282, 380, 438 rational equation, 132, 217 rational expression, 89, 100, 131, 1260, 1266 rational function, 585, 595, 601, 635 rational number, 131 rational numbers, 10, 15, 100 Rational Zero Theorem, 564, 635 ray, 810, 892 real number line, 13, 100 real numbers, 13, 100 reciprocal, 141, 363, 500 reciprocal function, 579 reciprocal identities, 979, 1049 reciprocal identity, 931, 940 rectangular coordinates, 1095, 1097, 1100 rectangular equation, 1104, 1158
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rectangular form, 1137, 1168 recursive formula, 1454, 1466, 1478, 1534 reduction formulas, 1012, 1049 reference angle, 822, 863, 877, 892 reflection, 676, 711 regression analysis, 774, 778, 781 regression line, 456 relation, 226, 380 remainder, 551 Remainder Theorem, 562, 635 removable discontinuity, 588, 635 Restricting the domain, 373 resultant, 1183, 1202 revenue function, 1225, 1331 Richter Scale, 685 This content is available for free at https://cnx.org/content/col11758/1.5 right triangle, 835, 950 roots, 477, 635 rose curve, 1126, 1202 row, 1272, 1331 row matrix, 1273 row operations, 1288, 1294, 1304, 1305, 1306, 1332 row-echelon form, 1288, 1292, 1332 row-equivalent, 1288, 1332 S sample space, 1522, 1534 SAS (side-angle-side) triangle, 1079 scalar, 1185, 1202, 1276 scalar multiple, 1186, 1276, 1332 Scalar multiplication, 1185 scalar multiplication, 1202, 1276 scatter plot, 451 scientific notation, 46, 48, 100 secant, 873, 892, 931 secant function, 931 sector of a circle, 827 sequence, 1446, 1463, 1534 series, 1487, 1534 set-builder notation, 204, 261, 380 sigma, 1487 sine, 977, 1023, 1025 sine function, 851, 892, 902, 907, 915, 919 sinusoidal function, 905, 967 slope, 136, 217, 395, 466 slope-intercept form, 395, 396, 401, 466 smooth curve, 516, 635 solution set, 129, 217, 1234, 1332 solving systems of linear equations, 1217 special angles, 989 square matrix, 1273, 1317 square root property, 178, 217 SSS (side-side-side) triangle, 1079 standard form, 139 standard form of a quadratic function, 481, 635 standard position, 811, 892, 1179, 1202 stretch
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