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Contents Lesson 11-1Arithmetic Sequences Lesson 11-2Arithmetic Series Lesson 11-3Geometric Sequences Lesson 11-4Geometric Series Lesson 11-5Infinite Geometric Series Lesson 11-6Recursion and Special Sequences Lesson 11-7The Binomial Theorem Lesson 11-8Proof and Mathematical Induction

Lesson 1 Contents Example 1Find the Next Terms Example 2Find a Particular Term Example 3Write an Equation for the nth Term Example 4Find Arithmetic Means

Example 1-1a Find the next four terms of the arithmetic sequence –8, –6, –4, …. Find the common difference d by subtracting 2 consecutive terms. Now add 2 to the third term of the sequence and then continue adding 2 until the next four terms are found. –4–2024 Answer: The next four terms are –2, 0, 2 and

Example 1-1b Find the next four terms of the arithmetic sequence 5, 3, 1, …. Answer: The next four terms are –1, –3, –5 and –7.

Example 1-2a Construction The table below shows typical costs for a construction company to rent a crane for one, two, three, or four months. Assuming that the arithmetic sequence continues, how much would it cost to rent the crane for 24 months? MonthsCost($) 175, , , ,000

Example 1-2a MonthsCost($) 175, , , ,000 Explore Since the difference between any two successive costs is $15,000, the costs form an arithmetic sequence with common difference 15,000. Plan You can use the formula for the nth term of an arithmetic sequence withand to find the cost for 24 months.

Example 1-2a Solve Formula for the nth term Simplify. Answer: It would cost $420,000 to rent for 24 months.

Example 1-2a Examine You can find the term of the sequence by adding 15,000. From Example 2 on page 579 of your textbook, you know the cost to rent the crane for 12 months is $120,000. So, a 12 through a 24 are 240,000, 255,000, 270,000, 285,000, 300,000, 315,000, 330,000, 345,000, 360,000, 375,000, 390,000, 405,000, and 420,000. Therefore, $420,000 is correct.

Example 1-2b Construction The table below shows typical costs for a construction company to rent a crane for one, two, three, or four months. Assuming that the arithmetic sequence continues, how much would it cost to rent the crane for 8 months? MonthsCost($) 175, , , ,000 Answer: It would cost $180,000 to rent for 8 months.

Example 1-3a Write an equation for the nth term of the arithmetic sequence –8, –6, –4, …. In this sequence,and Use the nth formula to write an equation. Formula for the nth term Distributive Property Simplify. Answer: An equation is.

Example 1-3b Write an equation for the nth term of the arithmetic sequence 5, 3, 1, …. Answer: An equation is.

Example 1-4a Find the three arithmetic means between 21 and 45. You can use the nth term formula to find the common difference. In the sequence 21, ___, ___, ___, 45,..., and Formula for the nth term Subtract 21 from each side. Divide each side by 4.

Example 1-4a Now use the value of d to find the three arithmetic means Answer: The arithmetic means are 27, 33, and 39. Check

Example 1-4b Find the three arithmetic means between 13 and 25. Answer: The arithmetic means are 16, 19, and 22.

End of Lesson 1

Lesson 2 Contents Example 1Find the Sum of an Arithmetic Series Example 2Find the First Term Example 3Find the First Three Terms Example 4Evaluate a Sum in Sigma Notation

Example 2-1a Find the sum of the first 20 even numbers, beginning with 2. The series is Since you can see that andyou can use either sum formula for this series. Method 1 Sum formula Multiply. Simplify.

Example 2-1a Method 2 Sum formula Multiply. Simplify. Answer: The sum of the first 20 even numbers is 420.

Example 2-1b Find the sum of the first 15 numbers, beginning with 1. Answer: The sum of the first 15 numbers is 120.

Example 2-2a Radio A radio station is giving away money every day in the month of September for a total of $124,000. They plan to increase the amount of money given away by $100 each day. How much should they give away on the first day of September, rounded to the nearest cent? You know the values of n, S n, and d. Use the sum formula that contains d.

Example 2-2a Sum formula Simplify. Distributive Property Subtract 43,500. Divide by 30. Answer: They should give away $ the first day.

Example 2-2b Games A television game show gives contestants a chance to win a total of $1,000,000 by answering 16 consecutive questions correctly. If the value of each question is increased by $5,000, how much is the first question worth? Answer: The first question is worth $25,000.

Example 2-3a Find the first four terms of an arithmetic series in which Step 1 Since you knowanduse

Example 2-3a Step 2 Find d. Step 3 Use d to determine

Example 2-3a Answer: The first four terms are 14, 17, 20, 23.

Example 2-3b Find the first three terms of an arithmetic series in whichand Answer: The first three terms are 11, 16, 21.

Example 2-4a Evaluate Method 1 Find the terms by replacing k with 3, 4, …, 10. Then add.

Example 2-4a Method 2 Since the sum is an arithmetic series, use the formulaThere are 8 terms, Answer: The sum of the series is 112.

Example 2-4b Evaluate Answer: 108

End of Lesson 2

Lesson 3 Contents Example 1Find the Next Term Example 2Find a Particular Term Example 3Write an Equation for the nth Term Example 4Find a Term Given the Fourth Term and the Ratio Example 5Find Geometric Means

Example 3-1a Multiple-Choice Test Item Find the missing term in the geometric sequence 324, 108, 36, 12, ___. A 972 B 4 C 0 D –12 Read the Test Item Sincethe sequence has the common ratio of

Example 3-1a Answer: B Solve the Test Item To find the missing term, multiply the last given term by

Example 3-1b Multiple-Choice Test Item Find the missing term in the geometric sequence 100, 50, 25, ___. A 200 B 0 C 12.5 D –12.5 Answer: C

Example 3-2a Find the sixth term of a geometric sequence for which and Formula for the nth term Multiply. Answer: The sixth term is 96.

Example 3-2b Answer: The fifth term is 96. Find the fifth term of a geometric sequence for which and

Example 3-3a Write an equation for the nth term of the geometric sequence 5, 10, 20, 40, …. Formula for the nth term Answer: An equation is

Example 3-3b Write an equation for the nth term of the geometric sequence 2, 6, 18, 54, …. Answer: An equation is.

Example 3-4a Find the seventh term of a geometric sequence for whichand First find the value of Formula for the nth term Divide by 4.

Example 3-4a Now find a 7. Formula for the nth term Answer: The seventh term is Use a calculator.

Example 3-4b Answer: The sixth term is 243. Find the sixth term of a geometric sequence for whichand

Example 3-5a Find three geometric means between 3.12 and Use the nth term formula to find the value of r. In the sequence 3.12, ___, ___, ___, 49.92, a 1 is 3.12 and a 5 is Formula for the nth term Divide by Take the fourth root of each side.

Example 3-5a Answer: The geometric means are 6.24, 12.48, and 24.96, or –6.24, 12.48, and – There are two possible common ratios, so there are two possible sets of geometric means. Use each value of r to find three geometric means.

Example 3-5b Find three geometric means between 12 and Answer:The geometric means are 6, 3, and 1.5, or –6, 3, and –1.5.

End of Lesson 3

Lesson 4 Contents Example 1Find the Sum of the First n Terms Example 2Evaluate a Sum Written in Sigma Notation Example 3Use the Alternate Formula for a Sum Example 4Find the First Term of a Series

Example 4-1a Genealogy How many direct ancestors would a person have after 8 generations? Counting two parents, four grandparents, eight great- grandparents, and so on gives you a geometric series with Answer: Going back 8 generations, a person would have 510 ancestors. Sum formula Use a calculator.

Example 4-1b Genealogy How many direct ancestors would a person have after 7 generations? Answer: Going back 7 generations, a person would have 254 ancestors.

Example 4-2a Evaluate Method 1 Find the terms by replacing n with 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Then add.

Example 4-2a Method 2 Since the sum is a geometric series, you can use the formula Sum formula

Example 4-2a Simplify. Answer: The sum of the series is 12, 285.

Example 4-2b Evaluate Answer: 200

Example 4-3a Find the sum of a geometric series for which Since you do not know the value of n, use the formula

Example 4-3a Alternate sum formula Simplify. Answer: The sum of the series is 6666.

Example 4-3b Find the sum of a geometric series for which Answer: The sum of the series is 74.4.

Example 4-4a Find a 1 in a geometric series for which and Answer: The first term of the series is 3. Sum formula Divide each side by –255. Subtract.

Example 4-4b Find a 1 in a geometric series for which and Answer: The first term of the series is 1.

End of Lesson 4

Lesson 5 Contents Example 1Sum of an Infinite Geometric Series Example 2Infinite Series in Sigma Notation Example 3Write a Repeating Decimal as a Fraction

Example 5-1a Answer: The sum does not exist. Find the sum of, if it exists. First, find the value of r to determine if the sum exists. the sum does not exist.

Example 5-1a Find the sum of, if it exists. the sum exists.

Example 5-1a Now use the formula for the sum of an infinite geometric series. Sum formula Simplify. Answer: The sum of the series is 2.

Example 5-1b Find the sum of each infinite geometric series, if it exists. a. b. Answer: 2 Answer: no sum

Example 5-2a Evaluate In this infinite geometric series, Sum formula Simplify.

Example 5-2a Answer: Thus,

Example 5-2b Answer: 3 Evaluate

Example 5-3a Writeas a fraction. Method 1 Write the repeating decimal as a sum.

Example 5-3a In this series, Sum formula

Example 5-3a Subtract. Simplify.

Example 5-3a Method 2 Label the given decimal. Repeating decimal Multiply each side by 100. Subtract the second equation from the third. Divide each side by 99. Answer: Thus,

Example 5-3b Writeas a fraction. Answer:

End of Lesson 5

Lesson 6 Contents Example 1Use a Recursive Formula Example 2Find and Use a Recursive Formula Example 3Iterate a Function

Example 6-1a Find the first five terms of the sequence in which and Recursive formula

Example 6-1a Recursive formula Answer: The first five terms of the sequence are 5, 17, 41, 89, 185.

Example 6-1b Find the first five terms of the sequence in which and Answer: The first five terms of the sequence are 2, 8, 26, 80, 242.

Example 6-2a Biology Dr. Elliott is growing cells in lab dishes. She starts with 108 cells Monday morning and then removes 20 of these for her experiment. By Tuesday the remaining cells have multiplied by 1.5. She again removes 20. This pattern repeats each day in the week. Write a recursive formula for the number of cells Dr. Elliott finds each day before she removes any. Let c n represent the number of cells at the beginning of the nth day. She takes 20 out, leaving c n – 20. The number the next day will be 1.5 times as much. So, Answer:

Example 6-2a Find the number of cells she will find on Friday morning. On the first morning, there were 108 cells, so Recursive formula

Example 6-2a Answer: On the fifth day, there will 303 cells. Recursive formula

Example 6-2b Answer: Biology Dr. Scott is growing cells in lab dishes. She starts with 100 cells Monday morning and then removes 30 of these for her experiment. By Tuesday the remaining cells have doubled. She again removes 30. This pattern repeats each day in the week. a. Write a recursive formula for the number of cells Dr. Scott finds each day before she removes any. b. Find the number of cells she will find on Saturday morning. Answer: 1340

Example 6-3a Find the first three iterates x 1, x 2, and x 3 of the function for an initial value of To find the first iterate x 1, find the value of the function for Iterate the function. Simplify.

Example 6-3a To find the second iterate x 2, substitute x 1 for x. Iterate the function. Simplify. Substitute x 2 for x to find the third iterate. Iterate the function. Simplify.

Example 6-3a Answer: Therefore, 5, 14, 41, 122 is an example of a sequence generated using iteration.

Example 6-3b Find the first three iterates x 1, x 2, and x 3 of the function for an initial value of Answer: 5, 11, 23

End of Lesson 6

Lesson 7 Contents Example 1Use Pascal’s Triangle Example 2Use the Binomial Theorem Example 3Factorials Example 4Use a Factorial Form of the Binomial Theorem Example 5Find a Particular Term

Example 7-1a Expand Write row 5 of Pascal’s triangle Use the patterns of a binomial expansion and the coefficients to write the expansion of Answer:

Example 7-1b Expand Answer:

The expression will have nine terms. Use the sequence to find the coefficients for the first five terms. Use symmetry to find the remaining coefficients. Example 7-2a Expand

Example 7-2a Answer:

Example 7-2b Expand Answer:

Example 7-3a Evaluate Answer: 1 1

Example 7-3b Evaluate Answer: 420

Example 7-4a Expand Let Binomial Theorem, factorial form

Example 7-4a Simplify.

Example 7-4a Answer:

Example 7-4b Expand Answer:

Example 7-5a Find the fourth term in the expansion of First, use the Binomial Theorem to write the expression in sigma notation. In the fourth term,

Example 7-5a Answer: Simplify.

Example 7-5b Find the fifth term in the expansion of Answer:

End of Lesson 7

Lesson 8 Contents Example 1Summation Formula Example 2Divisibility Example 3Counterexample

Example 8-1a Prove that Step 1 When, the left side of the given equation is 2(1) –1 or 1. The right side is 1 2 or 1. Thus, the equation is true for Step 2 Assume for a positive integer k. Step 3 Show that the given equation is true for

Example 8-1a Addto each side. Add.Simplify. Factor.

Example 8-1a The last expression is the right side of the equation to be proved, where n has been replaced byThus, the equation is true for Answer: This proves that is true for all positive integers n.

Example 8-1b Prove that. Step 1 When, we have, which is true. Thus, the equation is true for Answer: Step 2 Assume that for a positive integer k. Step 3 Show that the given equation is true for

Example 8-1b The last expression is the right side of the equation to be proved, where n has been replaced by Thus, the equation is true for This proves that is true for all positive integers n.

Example 8-2a Prove that is divisible by 5 for all positive integers n. Step 1 When, Since 5 is divisible by 5, the statement is true for Step 2 Assume thatis divisible by 5 for some positive integer k. This means that there is a whole number r such that

Example 8-2a Step 3 Show that the statement is true for Inductive hypothesis Add 1 to each side. Multiply each side by 6. Simplify. Subtract 1 from each side. Factor.

Example 8-2a Since r is a whole number,is a whole number. Thereforeis divisible by 5. Answer: Thus, the statement is true forThis proves that is divisible by 5.

Example 8-2b Prove that is divisible by 9 for all positive integers n. Step 1 When, Since 9 is divisible by 9, the statement is true for Step 2 Assume thatis divisible by 9 for some positive integer k. Answer:

Example 8-2b Step 3 Show that this is true for.

Example 8-2b Since r is a whole number,is a whole number. Therefore,is divisible by 9. Thus, the statement is true forThis proves that is divisible by 9.

Example 8-3a Find a counterexample for the formula that is always a prime number for any positive integer n. Check the first few positive integers. nFormulaprime? or 7yes or 11yes or 17yes or 25no Answer: The valueis a counterexample for the formula.

Example 8-3b Find a counterexample for the formula that is always a prime number for any positive integer n. Answer: The valueis a counterexample for the formula.

End of Lesson 8

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