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Calculus Level 4

$\large\displaystyle\sum_{k=1}^{\infty}\dfrac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}$

Find the value of the above series.

1 solution

Sudhamsh Suraj
Mar 12, 2017

write $6^k$ = $2^k(3.3^k-2.2^k) - 2.2^k(3^k-2^k)$

Now by using telescopic cancellations .

We get answer as 2 .

$note$ :

This is a converging series . So telescopic cancellation can be used and leaving last term as it tends to infinity