Chapter 11 - 11.1 SOLUTIONS 485 CHAPTER ELEVEN Solutions...

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Unformatted text preview: 11.1 SOLUTIONS 485 CHAPTER ELEVEN Solutions for Section 11.1 1. Adding the terms, we see that 3 + 3 2 + 3 2 2 = 3 + 6 + 12 = 21 . We can also find the sum using the formula for a finite geometric series with a = 3 , r = 2 , and n = 3 : 3 + 3 2 + 3 2 2 = 3(1- 2 3 ) 1- 2 = 3(8- 1) = 21 . 2. Adding the terms, we see that 50 + 50(0 . 9) + 50(0 . 9) 2 + 50(0 . 9) 3 = 50 + 45 + 40 . 5 + 36 . 45 = 171 . 95 . We can also find the sum using the formula for a finite geometric series with a = 50 , r = 0 . 9 , and n = 4 : 50 + 50(0 . 9) + 50(0 . 9) 2 + 50(0 . 9) 3 = 50(1- (0 . 9) 4 ) 1- . 9 = 171 . 95 . 3. The sum can be rewritten as 2 + 2(2) + 2(2 2 ) + + 2(2 9 ) . This is a finite geometric series with a = 2 , r = 2 , and n = 10 . We have Sum = 2(1- (2) 10 ) 1- 2 = 2046 . 4. We use the formula for the sum of a finite geometric series with a = 20 , r = 1 . 4 , and n = 9 . We have Sum = 20(1- (1 . 4) 9 ) 1- 1 . 4 = 983 . 05 . 5. This is an infinite geometric series with a = 1000 and r = 1 . 08 . Since r > 1 , the series diverges and the sum does not exist. 6. We use the formula for the sum of a finite geometric series with a = 500 , r = 0 . 6 , and n = 16 . We have Sum = 500(1- (0 . 6) 16 ) 1- . 6 = 1249 . 65 . 7. This is an infinite geometric series with a = 30 and r = 0 . 85 . Since- 1 < r < 1 , the series converges and we have Sum = 30 1- . 85 = 200 . 8. This is an infinite geometric series with a = 25 and r = 0 . 2 . Since- 1 < r < 1 , the series converges and we have Sum = 25 1- . 2 = 31 . 25 . 486 Chapter Eleven /SOLUTIONS 9. We use the formula for the sum of a finite geometric series with a = 1 , r = 1 / 2 , and n = 9 . We have Sum = 1- (0 . 5) 9 1- . 5 = 1 . 9961 . 10. This is an infinite geometric series with a = 1 and r = 1 / 3 . Since- 1 < r < 1 , the series converges and we have Sum = 1 1- 1 / 3 = 1 . 5 . 11. This is a finite geometric series with a = 3 , r = 1 / 2 , and n- 1 = 10 , so n = 11 . Thus Sum = 3(1- (1 / 2) 11 ) 1- 1 / 2 = 3 2 (2 11- 1) 2 11 = 3(2 11- 1) 2 10 . 12. Each term in this series is 1 . 5 times the preceding term, so this is an infinite geometric series with a = 1000 and r = 1 . 5 . Since r > 1 , the series diverges and the sum does not exist. 13. Each term in this series is half the preceding term, so this is an infinite geometric series with a = 200 and r = 0 . 5 . Since- 1 < r < 1 , the series converges and we have Sum = 200 1- . 5 = 400 . 14. This is an infinite geometric series with a =- 2 and r =- 1 / 2 . Since- 1 < r < 1 , the series converges and Sum =- 2 1- (- 1 / 2) =- 4 3 . 15. Since a = 10 and r = 0 . 75 , we find the partial sums using the formula S n = 10(1- (0 . 75) n ) 1- . 75 . For n = 5 , we have S 5 = 10(1- (0 . 75) 5 ) 1- . 75 = 30 . 51 ....
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This note was uploaded on 04/21/2008 for the course MATH 227 taught by Professor ? during the Winter '08 term at University of Cincinnati.

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Chapter 11 - 11.1 SOLUTIONS 485 CHAPTER ELEVEN Solutions...

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