This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Su, Yung Homework 11 Due: Nov 13 2007, 3:00 am Inst: Shinko Harper 1 This printout should have 22 questions. Multiplechoice 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...
View
Full
Document
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.
 Spring '09
 Harper
 Calculus

Click to edit the document details