q11-sol-1042-sp08

q11-sol-1042-sp08 - Calculus II-1042 Solution of Quiz 11...

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Calculus II—1042 Solution of Quiz # 11 Fri., 04/11/08 1. Determine if the Geometric series convergent or divergent. If convergent, find its sum. (a) X n =1 ( - 1) n (3) n - 1 4 n +1 . Solution: It is a Geometeric Series with starting point n = 1. First write the series in a correct form X n =1 ar n - 1 X n =1 ( - 1) n (3) n - 1 4 n +1 = X n =1 ( - 1)( - 1) n - 1 (3) n - 1 4 2 · 4 n - 1 = X n =1 - 1 16 - 3 4 n - 1 Now a = - 1 16 and r = - 3 4 so the Series is covergent as | r | = 3 4 < 1 and Sum = - 1 16 1 - - 3 4 = - 1 28 (b) X n =0 π n e n +1 . Solution: It is a Geometeric Series with starting point n = 0. First write the series in a correct form X n =0 ar n X n =0 π n e n +1 = X n =0 π n e · e n = X n =0 1 e π e · n a = 1 e and r = π e so the Series is divergent as | r | = π e > 1 2. Find the values of x for which the given Geometric series X n =0 4 x - 3 2 n converges. Find the sum of the series for those values of x . Solution: It is a Geometeric Series with starting point n = 0.
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