Sequence and series (7)

Algebra Level pending

∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) = ? \sum_{k=1}^\infty \frac{6^k}{(3^k - 2^k)(3^{k+1} - 2^{k+1})} =\ ?

2 2 3 3 1 2 \frac12 1 3 \frac13

Adhiraj Dutta by Adhiraj Dutta , 18, India

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1 solution

Chew-Seong Cheong
Apr 24, 2020

S = ∑ k = 1 ∞ 6 k ( 3 k − 2 k ) ( 3 k + 1 − 2 k + 1 ) By partial fraction decomposition = ∑ k = 1 ∞ ( 2 k 3 k − 2 k − 2 k + 1 3 k + 1 − 2 k + 1 ) = 2 3 − 2 = 2 \begin{aligned} S & = \sum_{k=1}^\infty \frac {6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})} & \small \blue{\text{By partial fraction decomposition}} \\ & = \sum_{k=1}^\infty \left(\frac {2^k}{3^k-2^k} - \frac {2^{k+1}}{3^{k+1}-2^{k+1}}\right) \\ & = \frac 2{3-2} = \boxed 2 \end{aligned}

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You can make this a cleanly telescoping sum if instead of 3 k 3^k and 3 k + 1 3^{k+1} in the numerators, you use 2 k 2^k and 2 k + 1 . 2^{k+1}.

Patrick Corn - 1 year, 1 month ago

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Thanks, I didn't see that.

Chew-Seong Cheong - 1 year, 1 month ago

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