Chapter 11 - 11.1 SOLUTIONS 485 CHAPTER ELEVEN Solutions for Section 11.1 1 Adding the terms we see that We can also find the sum using the formula for

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

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11.1 SOLUTIONS 485 CHAPTER ELEVEN Solutions for Section 11.1 1.Adding the terms, we see that 2.Adding the terms, we see that 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 - 0 . 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 - 0 . 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 - 0 . 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 - 0 . 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 - 0 . 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 - 0 . 75 . For n = 5 , we have S 5 = 10(1 - (0 . 75) 5 ) 1 - 0 . 75 = 30 . 51 . For n = 10 , we have S 10 = 10(1 - (0 . 75) 10 ) 1 - 0 . 75 = 37 . 75 . For n = 15 , we have S 15 = 10(1 - (0 . 75) 15 ) 1 - 0 . 75 = 39 . 47 . For n = 20 , we have S 20 = 10(1 - (0 . 75) 20 ) 1 - 0 . 75 = 39 . 87 . As n gets larger, the partial sums appear to be approaching 40 , as we expect. 16. We find the partial sums using the formula S n = 250(1 - (1 . 2) n ) 1 - 1 . 2 . For n = 5 , we have S 5 = 250(1 - (1 . 2) 5 ) 1 - 1 . 2 = 1 , 860 . 40 . For n = 10 , we have S 10 = 250(1 - (1 . 2) 10 ) 1 - 1 . 2 = 6 , 489 . 67 . For n = 15 , we have S 15 = 250(1 - (1 . 2) 15 ) 1 - 1 . 2 = 18 , 008 . 78 . For n = 20 , we have S 20 = 250(1 - (1 . 2) 20 ) 1 - 1 . 2 = 46 , 672 . 00 . As n gets larger, the partial sums appear to be growing without bound, as we expect, since r > 1 .
11.1 SOLUTIONS 487 17. (a) Notice that the 6 th deposit is made 5 months after the first deposit, so the first deposit has grown to 500(1 . 005) 5 at that time. The balance in the account right after the 6 th deposit is the sum Balance = 500 + 500(1 . 005) + 500(1 . 005) 2 + · · · + 500(1 . 005) 5 .
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