HW 11 key - Miller Kierste Homework 11 Due 3:00 am Inst...

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Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points Rewrite the finite sum 15 3 + 3 - 20 4 + 3 + 25 5 + 3 - 30 6 + 3 + . . . + ( - 1) 5 40 8 + 3 using summation notation. 1. 5 X k = 0 ( - 1) k - 3 5 k k + 3 2. 8 X k = 3 5 k k + 3 3. 8 X k = 3 ( - 1) k - 4 5 k k + 3 4. 9 X k = 4 ( - 1) k - 3 5 k k + 3 5. 5 X k = 0 5 k k + 3 6. 8 X k = 3 ( - 1) k - 3 5 k k + 3 correct Explanation: The numerators form a sequence 15 , 20 , 25 , 30 , . . . , 40 , while the denominators form a sequence 3 + 3 , 4 + 3 , 5 + 3 , 6 + 3 , . . . , 8 + 3 . Thus the general term in the series is of the form a k = ( - 1) k - 3 5 k k + 3 where the sum ranges from k = 3 to k = 8. Consequently, the series becomes 8 X k = 3 ( - 1) k - 3 5 k k + 3 in summation notation. keywords: finite sum, summation notation, 002 (part 1 of 1) 10 points Find the sum of the finite series 8 + 8 · 5 8 + 8 · 5 2 8 2 + . . . + 8 · 5 8 8 8 . 1. sum = 8 8 8 8 7 - 5 7 3 · 2. sum = 8 8 8 8 8 - 5 8 3 · 3. sum = 8 8 8 8 9 - 5 9 3 · correct 4. sum = 8 8 9 - 5 9 3 · 5. sum = 8 8 8 - 5 8 3 · Explanation: The given series is a finite geometric series 8 X n = 0 ar n , with a = 8 , r = 5 8 . Now 8 X n = 0 ar n = a 1 - r 9 1 - r · . Consequently, sum = 8 8 8 8 9 - 5 9 3 · . keywords: finite geometric series
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Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 2 003 (part 1 of 3) 10 points If the n th partial sum of n = 1 a n is S n = 2 n - 5 n + 1 , (i) what is a 1 ? 1. a 1 = 7 2 2. a 1 = - 3 2 correct 3. a 1 = 3 2 4. a 1 = 2 5. a 1 = - 7 2 Explanation: Since a 1 = S 1 , a 1 = - 3 2 . 004 (part 2 of 3) 10 points (ii) What is a n for n > 1? 1. a n = 3 n 2 2. a n = 7 n 2 3. a n = 3 n ( n - 1) 4. a n = 7 n ( n + 1) correct 5. a n = 7 n ( n - 1) 6. a n = 3 n ( n + 1) Explanation: Since S n = a 1 + a 2 + · · · + a n , we see that a n = S n - S n - 1 . But S n = 2 n - 5 n + 1 = 2( n + 1) - 7 n + 1 = 2 - 7 n + 1 . Consequently, a n = 7 n - 7 n + 1 = 7 n ( n + 1) for all n > 1. 005 (part 3 of 3) 10 points (iii) What is the sum n = 1 a n ? 1. sum = - 1 2. sum = 1 3. sum = 2 correct 4. sum = 0 5. sum = - 2 Explanation: By definition sum = lim n →∞ S n = lim n → ∞ 2 n - 5 n + 1 · . Thus sum = 2 . keywords: 006 (part 1 of 1) 10 points Determine whether the series X n = 0 4 (cos ) 1 2 n is convergent or divergent, and if convergent, find its sum. 1. convergent with sum 8
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Miller, Kierste – Homework 11 – Due: Apr 10 2007, 3:00 am – Inst: Gary Berg 3 2. divergent 3. convergent with sum - 3 8 4. convergent with sum 8 3 correct 5. convergent with sum - 8 3 6. convergent with sum - 8 Explanation: Since cos = ( - 1) n , the given series can be rewritten as an infinite geometric series X n =0 4 - 1 2 n = X n = 0 a r n in which a = 4 , r = - 1 2 . But the series n =0 ar n is (i) convergent with sum a 1 - r when | r | < 1, and (ii) divergent when | r | ≥ 1. Consequently, the given series is convergent with sum 8 3 . keywords: geometric series, convergent 007 (part 1 of 1) 10 points Determine whether the series X n = 1 n 2 5 n 2 + 4 is convergent or divergent, and if convergent, find its sum.
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