a. Explanation Calculation: Consider the series, ∑ k = 1 ∞ 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) (1) Obtain the partial fraction of 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) . 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) = a 0 3 k + 1 − 1 + a 1 3 k − 1 (2) Multiply the equation by ( 3 k + 1 − 1 ) ( 3 k − 1 ) and simplify the terms, 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) ( 3 k + 1 − 1 ) ( 3 k − 1 ) = a 0 ( 3 k + 1 − 1 ) ( 3 k − 1 ) ( 3 k − 1 ) + a 1 ( 3 k + 1 − 1 ) ( 3 k − 1 ) 3 k + 1 − 1 3 k = a 0 ( 3 k + 1 − 1 ) + a 1 ( 3 k − 1 ) 3 k = 3 k + 1 a 0 − a 0 + 3 k a 1 − a 1 3 k = 3 ( a 0 + a 1 ) 3 k − ( a 0 + a 1 ) Equate the coefficient of 3 k and the constant term on both sides, 3 ( a 0 + a 1 ) = 1 a 0 + a 1 = 0 Solve the above equations and obtain a 0 = − 1 2 and a 1 = 1 2 . Substitute − 1 2 for a 0 and 1 2 for a 1 in equation (3), 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) = − 1 2 3 k + 1 − 1 + 1 2 3 k − 1 Therefore, the partial fraction is 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) = 1 2 ( 3 k − 1 ) − 1 2 ( 3 k + 1 − 1 ) (3) Substitute equation (3) in equation (1), the series is ∑ k = 1 ∞ 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) = ∑ k = 1 ∞ ( 1 2 ( 3 k − 1 ) − 1 2 ( 3 k + 1 − 1 ) ) b. To determine To find: The value of a which the series ∑ k = 1 ∞ a k ( a k + 1 − 1 ) ( a k − 1 ) converge.

Calculus Early Transcendentals 3rd.edition I.r.c.

3rd Edition

ISBN: 9780134766843

Author: Briggs

Publisher: PEARSON

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Chapter 10.3, Problem 92E

To determine

To evaluate: The series ∑k=1∞3k(3k+1−1)(3k−1).

To determine

To find: The value of a which the series ∑k=1∞ak(ak+1−1)(ak−1) converge.

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Chapter 10.3, Problem 91E

Chapter 10.3, Problem 93E

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The series ∑ k = 1 ∞ 3 k ( 3 k + 1 − 1 ) ( 3 k − 1 ) .

Solution Summary: The author explains how to calculate the sum of the series 14.