1. Copyright © 2004 Pearson Education, Inc.

2. Chapter 7 Further Topics in Algebra

3. Copyright © 2004 Pearson Education, Inc. 7.1 Sequences and Series

4. Slide 7-4 Copyright © 2004 Pearson Education, Inc. Sequences A sequence is a function that computes an ordered list. Instead of using f(x) notation to indicate a sequence, it is customary to use a n, where a n = f(n). The letter n is used instead of x as a reminder that n represents a natural number. The elements in the range of a sequence, called the terms of the sequence, are a 1, a 2, a 3,…. The elements of both the domain and the range of a sequence are ordered. The first term is found by letting n = 1, the second terms is found by letting n = 2, and so on. The general term, or nth term, of the sequence is a n.

5. Slide 7-5 Copyright © 2004 Pearson Education, Inc. Examples a n = 3n  5 Solution: Replacing n in a n = 3n  5, with 1, 2, 3, and 4 gives n = 1: a 1 = 3(1)  5 =  2 n = 2: a 2 = 3(2)  5 = 1 n = 3: a 3 = 3(3)  5 = 4 n = 4: a 4 = 3(4)  5 = 7 Solution: Replacing n in, with 1, 2, 3, and 4 gives n = 1: a 1 = n = 2: a 2 = n = 3: a 3 = n = 4: a 4 = Find the first four terms for each sequence.

6. Slide 7-6 Copyright © 2004 Pearson Education, Inc. Sequences A sequence is a finite sequence if the domain is the set{1, 2, 3, …n}, where n is a natural number. An infinite sequence has the set of all natural numbers as its domain.  For example, the sequence of natural-number multiples of 2, 2, 4, 6, 8, 10, 12, 14,…, is infinite, but the sequence of days in June, 1, 2, 3, 4, …., 29, 30, is finite.

7. Slide 7-7 Copyright © 2004 Pearson Education, Inc. Recursion Formula Some sequences are defined by recursive definition, one which each term after the first term or first few terms is defined as an expression involving the previous term or terms. Example: a 1 = 3 Solution: This is a recursive definition. We know a 1 = 3. Since

8. Slide 7-8 Copyright © 2004 Pearson Education, Inc. Series and Summation Notation The sum of a sequence, called a series, is written using summation notation. Series A finite series is an expression of the form And an infinite series is an expression of the form The letter i is called the index of the summation.

9. Slide 7-9 Copyright © 2004 Pearson Education, Inc. Example Evaluate the series Solution: Write each of the four terms, then evaluate the sum.

10. Slide 7-10 Copyright © 2004 Pearson Education, Inc. Summation Properties

11. Slide 7-11 Copyright © 2004 Pearson Education, Inc. Summation Rules

12. Slide 7-12 Copyright © 2004 Pearson Education, Inc. Summation Properties Example: Solution: Example: Solution: Use the summation properties to find each sum.

13. Slide 7-13 Copyright © 2004 Pearson Education, Inc. Using the Summation Properties Example: Evaluate Solution:

14. Copyright © 2004 Pearson Education, Inc. 7.2 Arithmetic Sequences and Series

15. Slide 7-15 Copyright © 2004 Pearson Education, Inc. Arithmetic Sequences A sequence in which each term after the first is obtained by adding a fixed number to the previous term is an arithmetic sequence. The fixed number that is added is the common difference. The sequence 7, 10, 13, 16, 19,… is an arithmetic sequence since each term after the first is obtained by adding 3 to the previous. The common difference is 3.

16. Slide 7-16 Copyright © 2004 Pearson Education, Inc. Common Difference The common difference of an arithmetic sequence is d. Then by definition of an arithmetic sequence, d = a n+ 1  a n, for every positive integer n in its domain. Example: Find the common difference, d, for the arithmetic sequence  11,  7,  3, 1, 5,…. Solution: We find d by choosing any two adjacent terms and subtracting the first from the second. Choosing  7 and  3 gives d =  3  (  7) = 4 Choosing 1 and 5 would give d = 5  1 =4, the same result.

17. Slide 7-17 Copyright © 2004 Pearson Education, Inc. Example Write the first five terms for each arithmetic sequence. The first term is 5, and the common difference is  2. Solution: Here a 1 = 5 and d =  2, so a 2 = 5 + (  2) = 3 a 3 = 3 + (  2) = 1 a 4 = 1 + (  2) =  1 a 5 =  1 + (  2) =  3

18. Slide 7-18 Copyright © 2004 Pearson Education, Inc. Example continued a 1 =  8, d = 6 Solution: Starting with a 1, add d to each term to get the next term. a 1 =  8 a 2 =  8 + d =  8 + 6 =  2 a 3 =  2 + d =  2 + 6 = 4 a 4 = 4 + d = 4 + 6 = 10 a 5 = 10 + d = 10 + 6 = 16

19. Slide 7-19 Copyright © 2004 Pearson Education, Inc. nth Term of a Arithmetic Sequence In an arithmetic sequence with first term a 1 and common difference d, the nth term, a n, is given by a n = a 1 + (n  1)d. Example: Find a 17 and a n for the arithmetic sequence  5,  2, 1, 4,….

20. Slide 7-20 Copyright © 2004 Pearson Education, Inc. nth Term of a Arithmetic Sequence continued Solution: Here a 1 =  5 and d = 1  (  2) = 3. To find a 17, substitute 17 for n in the formula for the nth term. a n = a 1 + (n  1)d a 17 = a 1 + (17  1)d a 17 =  5 + (16)3 a 17 = 43 Find a n by substituting values for a 1 and d in the formula for a n. a n =  5 + (n  1)  3 a n =  5 + 3n  3 a n = 3n  8

21. Slide 7-21 Copyright © 2004 Pearson Education, Inc. Another Example Suppose that an arithmetic sequence has a 3 =  5 and a 9 = 37. Find a 1. Solution: Since a 9 =a 3 + 6d, it follows that 6d = a 9  a 3 = 37  (  5) = 42 And so d = 7. To find a 1, use the equation a 3 = a 1 + 2d,  5 = a 1 + 2d  5 = a 1 + 2(7) a 1 =  19

22. Slide 7-22 Copyright © 2004 Pearson Education, Inc. Sum of the First n Terms of an Arithmetic Series If an arithmetic sequence has first term a 1 and common difference d, then the sum of the first n terms is given by Example: Evaluate S 8 for the arithmetic sequence  12,  7,  2, 3, 8,…. Solution: We want the sum of the first 8 terms. Using a 1 =  12, n = 8, and d = 5 in the second formula.

23. Slide 7-23 Copyright © 2004 Pearson Education, Inc. Summation Notation Any sum of the form where d and p are real numbers, represents the sum of the terms of an arithmetic sequence having first terms a 1 = d(1) + p = d + p and common difference d. These sums can be evaluated using the formulas in this section.

24. Slide 7-24 Copyright © 2004 Pearson Education, Inc. Examples Evaluate each sum. Solution: This sum contains the first 12 terms of the arithmetic sequence having a 1 = 3  1 + 6 = 9 and a 12 = 3  12 + 6 = 42. Thus,

25. Slide 7-25 Copyright © 2004 Pearson Education, Inc. Examples continued Solution: The first few terms are [6  4(5)] + [6  4(6)] + [6  4(7)] +    =  14 + (  18) + (  22) +   . Thus, a 1 =  14 and d =  4. If the sequence started with k = 1, there would be 11 terms. Since it starts with 5, four of those terms are missing, so there are six terms and n = 6.

26. Copyright © 2004 Pearson Education, Inc. 7.3 Geometric Sequences and Series

27. Slide 7-27 Copyright © 2004 Pearson Education, Inc. Geometric Sequences A geometric sequence is a sequence in which each term after the first is obtained by multiplying the preceding term by a constant nonzero real number, called the common ratio. If the common ratio of a geometric sequence is r, then by the definition of a geometric sequence for every positive integer n. Therefore, we can find the common ratio by choosing any term except the first and dividing it by the preceding term.

28. Slide 7-28 Copyright © 2004 Pearson Education, Inc. Geometric Sequence nth Term of a Geometric Sequences In the geometric sequence with first term a 1 and common ratio r, the nth term is a n = a 1 r n  1.

29. Slide 7-29 Copyright © 2004 Pearson Education, Inc. Finding Terms of a Geometric Sequence Example: Find a 7 and a n for the geometric sequence 2, 8, 32, 128,… Solution: The first term, a 1 is 2. Find r by choosing any term except the first and dividing it by the preceding term. We can find the seventh term using the formula for a n, a n = a 1 r n  1, by replacing n with 7, r with 4 and a 1 with 2. a 7 = 2  (4) 7  1 = 2  4 6 = 8192 By the formula, a n = 2  4 n  1.

30. Slide 7-30 Copyright © 2004 Pearson Education, Inc. Another Example Find a 1 and r for the geometric sequence with fourth term  54 and sixth term  486. Solution: Use the formula for the nth term of a geometric sequence. n = 4, a 4 = a 1 r 3 =  54 n = 6, a 6 = a 1 r 5 =  486 Since a 1 r 3 =  54 Since a 1 r 3 =  54 and r = ±3, it follows that a 1 =  2 or 2.

31. Slide 7-31 Copyright © 2004 Pearson Education, Inc. Geometric Series A geometric series is the sum of the terms of a geometric sequence. Sum of the First n Terms of a Geometric Sequence If a geometric sequence has first term a 1 and common ratio r, then the corresponding geometric series is given by

32. Slide 7-32 Copyright © 2004 Pearson Education, Inc. Example Find Solution: This series is the sum of the first four terms of a geometric sequence having a 1 = 5  2 1 = 10 and r = 2.

33. Slide 7-33 Copyright © 2004 Pearson Education, Inc. Infinite Geometric Sequence This quotient, is called the sum of the terms of an infinite geometric sequence. Sum of the Terms of an Infinite Geometric Sequence The sum of an infinite geometric sequence with first term a 1 and common ratio r, where  1 < r < 1, is given by

34. Slide 7-34 Copyright © 2004 Pearson Education, Inc. Examples Solution: Here a 1 = r =. Since  1 < r < 1, the preceding formula applies Solution: Find each sum.

35. Slide 7-35 Copyright © 2004 Pearson Education, Inc. Annuities A sequence of equal payments made at equal periods of time, such as car payments, is called an annuity. If the payments are accumulated in an account (with no withdrawals), the sum of the payments and interest on the payments is called the future value of the annuity.

36. Slide 7-36 Copyright © 2004 Pearson Education, Inc. Example To save money for college, Matthew’s grandparents deposited $1500 at the end of each year for 6 years in an account paying 4% interest, compounded annually. Find the future value of this annuity. Solution: To find the future value, we use the formula for compound interest A = P(1 + r) t. The first payment earns interest for 5 yr, the second payment for 4 yr, the third payment for 3 yr, the fourth payment for 2 yr, the fifth payment for 1 yr. The last payment earns no interest. The total amount is 1500(1.04) 5 + 1500(1.04) 4 + 1500(1.04) 3 + 1500(1.04) 2 + 1500(1.04) + 1500.

37. Slide 7-37 Copyright © 2004 Pearson Education, Inc. Example continued This is the sum of the terms of a geometric sequence with first term a 1 = 1500 and common ratio r = 1.04. Using the formula for S 6, the sum of six terms, gives The future value of the annuity is $9949.46

38. Slide 7-38 Copyright © 2004 Pearson Education, Inc. Future Value of an Annuity The formula for the future value of an annuity is given by Where S is future value, R is payment at the end of each period, i is interest rate per period, and n is number of periods.

39. Copyright © 2004 Pearson Education, Inc. 7.4 The Binomial Theorem

40. Slide 7-40 Copyright © 2004 Pearson Education, Inc. Pascal’s Triangle

41. Slide 7-41 Copyright © 2004 Pearson Education, Inc. n-Factorial For any positive integer n, n! = n(n  1)(n  2) …(3)(2)(1) and 0! = 1. Binomial Coefficient For nonnegative integers n and r, with r  n,

42. Slide 7-42 Copyright © 2004 Pearson Education, Inc. Example Evaluate each binomial coefficient. a) b)c)d) a)b) c)d)

43. Slide 7-43 Copyright © 2004 Pearson Education, Inc. The Binomial Theorem There are n + 1 terms The first term is x n, and the last term is y n. In each succeeding term, the exponent on x decreases by 1 and the exponent on y increases by 1. The sum of the exponents on the x and y in any term is n. The coefficient of the term with x r y n  r or x n  r y r is

44. Slide 7-44 Copyright © 2004 Pearson Education, Inc. Binomial Theorem For any positive integer n and any complex numbers x and y.

45. Slide 7-45 Copyright © 2004 Pearson Education, Inc. Example Write the binomial expansion of (x + y) 4. Using the binomial theorem. Now evaluate each of the binomial coefficients.

46. Slide 7-46 Copyright © 2004 Pearson Education, Inc. Example Expand (x  3y) 4

47. Slide 7-47 Copyright © 2004 Pearson Education, Inc. kth Term of the Binomial Expansion The kth term of the binomial expansion of (x + y) n, where n  k  1, is

48. Slide 7-48 Copyright © 2004 Pearson Education, Inc. Example Find the seventh term in the expansion (x 2  2y) 11. In the seventh term, that x = x 2, y =  2y, and n = 11. Then the 7 th term of the expansion is

49. Copyright © 2004 Pearson Education, Inc. 7.5 Mathematical Induction

50. Slide 7-50 Copyright © 2004 Pearson Education, Inc. Principle of Mathematical Induction Let S n be a statement concerning the positive integer n. Suppose that  S 1 is true;  for any positive integer k, k  n, if S k is true, then S k+1 is also true. Then S n is true for every positive integer value of n.

51. Slide 7-51 Copyright © 2004 Pearson Education, Inc. Proof by Mathematical Induction Step 1  Prove that the statement is true for n = 1. Step 2  Show that, for any positive integer k, k  n, if S k is true, then S k+1 is also true.

52. Slide 7-52 Copyright © 2004 Pearson Education, Inc. Example Let S n represent the statement Step 1: Show that the statement is true when n = 1. If n = 1, S 1 becomes

53. Slide 7-53 Copyright © 2004 Pearson Education, Inc. Example continued Step 2 Show that S k implies S k+1 is the statement and S k+1 is the statement Start with s k and assume it is a true statement.

54. Slide 7-54 Copyright © 2004 Pearson Education, Inc. Example continued Add k + 1 to both sides of this equation to obtain S k+1. This final result is the statement for n = k + 1; it has been shown that if S k is true, then S k+1 is also true.

55. Slide 7-55 Copyright © 2004 Pearson Education, Inc. Proving an Inequality Statement Prove: If x is a real number between 0 and 1, then for every positive integer n, 0 < x n < 1. Step 1Here S 1 is the statement if 0 < x < 1, then 0 < x 1 < 1, which is true. Step 2S k is the statement if 0 < x < 1, then 0 < x k < 1. To show that this implies that S k+1 is true, multiply all three parts by x.

56. Slide 7-56 Copyright © 2004 Pearson Education, Inc. Proving an Inequality Statement continued x 0 < x x k < x 1 (Here the fact that 0 < x is used) Simplify to obtain 0 < x k+1 < x. Since x < 1, 0 < x k+1 < x < 1 and thus 0 < x k+1 < 1. This work shows that if S k is true, then S k+1 is true.

57. Copyright © 2004 Pearson Education, Inc. 7.6 Counting Theory

58. Slide 7-58 Copyright © 2004 Pearson Education, Inc. Fundamental Principle of Counting

59. Slide 7-59 Copyright © 2004 Pearson Education, Inc. Example A restaurant offers a choice of 3 types of eggs, 4 different meats, and 2 types of potatoes. Use the fundamental principle of counting to find the number of different eggs, meat, and potato choices that can be selected. Three events are involved: selecting an egg, meat and potato. egg choicesmeat choicespotato choices

60. Slide 7-60 Copyright © 2004 Pearson Education, Inc. Example A child has 8 different books he wants to arrange in a row. How many different arrangements are possible? 8 events are involved; selecting a book for the first spot, the second spot and so on. For the first spot the child has 8 choices, the child has 7 choices for the second spot, and so on.

61. Slide 7-61 Copyright © 2004 Pearson Education, Inc. Example Suppose the child in the previous example only wanted to arrange 4 of the 8 books in a row. How many arrangements of 4 books are possible? The child still has 8 ways to choose the fill the first spot, 7 ways for the second spot, 6 ways for the third spot, and 5 ways for the fourth spot. There are only 4 spots to be filled instead of 8. 8 7 6 5 = 1680 arrangements

62. Slide 7-62 Copyright © 2004 Pearson Education, Inc. Permutations of n Elements Taken r at a Time If P(n, r) denotes the number of permutations of n elements taken r at a time, with r  n, then

63. Slide 7-63 Copyright © 2004 Pearson Education, Inc. Example—Find each permutation a) The number of permutations of the letters L, M, and N Formula: As shown in the tree diagram the 6 permutations are LMN, LNM, MLN, MNL, NLM, NML

64. Slide 7-64 Copyright © 2004 Pearson Education, Inc. Example—Find each permutation continued b) The number of permutations of 2 of the letters L, M, and N Find P(3, 2) This result is the same as the answer in part (a). After the first two choices are made, the third is already determined since only one letter is left.

65. Slide 7-65 Copyright © 2004 Pearson Education, Inc. Example In how many ways can 5 students be seated in a row of 5 desks? Use P(n, r) with n = 5 and r = 5 to get

66. Slide 7-66 Copyright © 2004 Pearson Education, Inc. Combinations of n Elements Taken r at a Time If C(n, r) represents the number of combinations of n elements taken r at a time, with r  n, then

67. Slide 7-67 Copyright © 2004 Pearson Education, Inc. Example How many different committees of 4 people can be chosen from a group of 9? Here we wish to know the number of 4-element combinations that can be formed from a set of 9. (Order does not matter so we want combinations, not permutations.) There are 126 ways to select the committee.

68. Slide 7-68 Copyright © 2004 Pearson Education, Inc. Example Four math teachers are to be selected from a group of 20 to work on a special project. a) In how many different ways can the teacher be selected? b) In how many ways can the group of 4 be selected if a particular teacher must work on the project?

69. Slide 7-69 Copyright © 2004 Pearson Education, Inc. Solution a) We want combinations, not permutations, since order within the group does not matter. There are 4845 ways to select the group. b) Since 1 teacher is already selected, the problem is reduced to selecting 3 more from the remaining 19 teachers. In this case, the group can be selected 969 ways.

70. Slide 7-70 Copyright © 2004 Pearson Education, Inc. Distinguishing Between Permutations and Combinations Clue words: group, committee, sample, selection Clue words: arrangement, schedule, order Subsets of r items from a set of n items Arrangements of r items from a set of n items Order is not importantOrder is important. Repetitions are not allowed Number of ways of selecting r items out of n items CombinationsPermutations

71. Slide 7-71 Copyright © 2004 Pearson Education, Inc. Should permutations or combinations be used to solve each problem? How many ways can a landscaper plant 6 different trees? Since changing the order of the 6 trees results in a different arrangement, permutations should be used. A sample of 10 seeds is randomly selected from a shipment of 1000. How many different samples are possible? The order in which the 10 seeds are selected is not important. So combinations should be used.

72. Copyright © 2004 Pearson Education, Inc. 7.7 Basics of Probability

73. Slide 7-73 Copyright © 2004 Pearson Education, Inc. Definitions Sample Space: Set of all possible outcomes of a given experiment. Event: A subset of the sample space.

74. Slide 7-74 Copyright © 2004 Pearson Education, Inc. Probability of Event E In a sample space with equally likely outcomes, the probability of an event E, written P(E), is the ratio of the number of outcomes in sample space S that belong to event E, n(E), to the total number of outcomes in sample space S, n(S). That is,

75. Slide 7-75 Copyright © 2004 Pearson Education, Inc. Example The spinner shown at the bottom is spun. Write each event in set notation and give the probability of the event. a) E 1 : the number spun is even b) E 2 : the number spun is greater than 9 c) E 3 : the number spun is less than 10 d) E 4 : the number spun is 12 The sample space, S = {6, 7, 8, 9, 10, 11} 67 11 109 8

76. Slide 7-76 Copyright © 2004 Pearson Education, Inc. Solutions a) Since E 1 = {6, 8, 10} and b) E 2 = {10, 11} and c) E 3 = {6, 7, 8, 9} and d) E 4 =  and

77. Slide 7-77 Copyright © 2004 Pearson Education, Inc. Complements The set of all outcomes in a sample space that do not belong to event E is called the complement of E written In the experiment of drawing a card from a well shuffled-deck, find the probability of event E, the card is a 10, and event E '.

78. Slide 7-78 Copyright © 2004 Pearson Education, Inc. Odds The odds in favor of an event E are expressed as the ratio of P(E) to P(E').

79. Slide 7-79 Copyright © 2004 Pearson Education, Inc. Example A gumball is selected at random from a bowl containing 6 red gumballs and 8 of some other color. Find the odds in favor of red gumball being selected. Let E represent “a red gumball is selected.” The odds in favor are:

80. Slide 7-80 Copyright © 2004 Pearson Education, Inc. Probability of the Union of Two Events Mutually exclusive events are events that cannot occur simultaneously. For any events E and F,

81. Slide 7-81 Copyright © 2004 Pearson Education, Inc. Example One card is drawn from a well-shuffled deck of 52 cards. What is the probability of the following outcomes? a) The card is a diamond or a 2. The events are not mutually exclusive since it is possible to draw a two of diamonds. P(diamond or 2) =

82. Slide 7-82 Copyright © 2004 Pearson Education, Inc. Example continued b) The card is a 4 or a queen. The events are mutually exclusive because it is impossible to draw one card that is both a 4 and a queen. P(4 or Q) =

83. Slide 7-83 Copyright © 2004 Pearson Education, Inc. Properties of Probabilities For events E and F: 1. 0  P(E)  1 2. P(a certain event) = 1 3. P(an impossible event) = 0 4. 5.

84. Slide 7-84 Copyright © 2004 Pearson Education, Inc. Binomial Probability A binomial experiment is an experiment that consists of repeated independent trials with only two outcomes in each trial, success or failure.

85. Slide 7-85 Copyright © 2004 Pearson Education, Inc. Example An experiment consists of rolling a die 10 times. Find each probability. a) The probability that in exactly 4 of the rolls, the result is a 3. The probability p of a 3 on one roll is 1/6. Here n = 10 and r = 4 so

86. Slide 7-86 Copyright © 2004 Pearson Education, Inc. Example continued b) The probability that in exactly 9 of the rolls, the result is not a 3. n = 10 and r = 9 and p = 5/6