∞ 6^k ∑ ----

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OMG! You have badly mangled this one up. I recognized it only because I had come across it earlier.

You are looking for:

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

Verify that you can write this as

$\sum_{k=1}^{\infty}\left[ \frac{3^k}{(3^k-2^k)} - \frac{3^{k+1}}{(3^{k+1}-2^{k+1})} \right]$

which is a telescopic summation. I am sure this can completed now

Waiting for some quality posts in this site

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