Homework 11 - Su, Yung Homework 11 Due: Nov 13 2007, 3:00...

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Unformatted text preview: Su, Yung Homework 11 Due: Nov 13 2007, 3:00 am Inst: Shinko Harper 1 This print-out should have 22 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 If the n th partial sum of an infinite series is S n = 5 n 2- 2 2 n 2 + 3 , what is the sum of the series? 1. sum = 9 4 2. sum = 11 4 3. sum = 21 8 4. sum = 5 2 correct 5. sum = 19 8 Explanation: By definition sum = lim n S n = lim n 5 n 2- 2 2 n 2 + 3 . Thus sum = 5 2 . keywords: partial sum, definition of series 002 (part 1 of 1) 10 points If the n th partial sum of n =1 a n is given by S n = 3 n + 4 n + 4 , what is a n when n 2? 1. a n = 16 n ( n + 4) 2. a n = 8 n ( n + 4) 3. a n = 8 ( n + 4)( n + 3) correct 4. a n = 16 ( n + 4)( n + 3) 5. a n = 16 ( n + 4)( n + 5) 6. a n = 8 ( n + 4)( n + 5) Explanation: By definition S n = n X k 1 a n = a 1 + a 2 + ... + a n . Thus, for n 2, a n = S n- S n- 1 = 3 n + 4 n + 4- 3( n- 1) + 4 ( n- 1) + 4 . Consequently, a n = 8 ( n + 4)( n + 3) . keywords: partial sum, definition of series 003 (part 1 of 1) 10 points Determine whether the series 4- 16 3 + 64 9- 256 27 + is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 4 2. convergent with sum = 8 7 3. series is divergent correct 4. convergent with sum = 3 Su, Yung Homework 11 Due: Nov 13 2007, 3:00 am Inst: Shinko Harper 2 5. convergent with sum = 9 7 Explanation: The infinite series 4- 16 3 + 64 9- 256 27 + = X n = 1 ar n- 1 is an infinite geometric series with a = 4 , r =- 4 3 . But an infinite geometric series n = 1 ar n- 1 (i) converges when | r | < 1 and has sum = a 1- r while it (ii) diverges when | r | 1 . Consequently, the given series is divergent . keywords: infinite series, geometric series, di- vergent 004 (part 1 of 1) 10 points Determine whether the series X n = 1 2 n 5 n + 3 is convergent or divergent, and if convergent, find its sum. 1. convergent with sum = 1 4 2. convergent with sum = 4 3. convergent with sum = 5 2 4. convergent with sum = 2 5 5. divergent correct Explanation: The infinite series X n =1 a n is divergent when lim n a n exists but lim n a n 6 = 0 . Note for the given series, a n = 2 n 5 n + 3 = 2 5 + 3 n , so lim n a n = lim n 2 n 5 n + 3 = 2 5 6 = 0 . Thus the given series is divergent . keywords: 005 (part 1 of 1) 10 points Determine whether the series X n = 0 2 1 3 n is convergent or divergent, and if convergent, find its sum. 1. convergent, sum =- 7 2 2. convergent, sum = 7 2 3. divergent 4. convergent, sum = 3 2 5. convergent, sum = 3 correct Su, Yung Homework 11 Due: Nov 13 2007, 3:00 am Inst: Shinko Harper 3 Explanation: The given series is an infinite geometric series X n = 0 ar n with a = 2 and r = 1 3 . But the sum of such a series is (i) convergent with sum a 1...
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This note was uploaded on 04/15/2009 for the course M 59685 taught by Professor Harper during the Spring '09 term at University of Texas at Austin.

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Homework 11 - Su, Yung Homework 11 Due: Nov 13 2007, 3:00...

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