Find the sum $\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$ I need to find the sum,
$$\sum_{k=1}^\infty {\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)}}$$
I have tried to break the terms into partial fractions (method of differences) but am not able to do so. How to proceed?

First we can try to split things into two pieces:
$$\frac{6^k}{\left(3^k-2^k\right) \left(3^{k+1}-2^{k+1}\right)} = \dfrac{A}{3^k-2^k} + \dfrac{B}{3^{k+1}-2^{k+1}}$$
So we have $A \cdot (3^{k+1}-2^{k+1}) + B \cdot (3^k-2^k) = 6^k$ which can be arranged to $3^k (3 A + B) - 2^k (2 A + B) = 6^k$. 
If we make $2A+B=0$ and $3A+B=2^k$, this equality will hold. This suggests that $B=-2A$ and we also see that $3A-2A=A=2^k$. Therefore $B=-2^{k+1}$.
$$S = \sum_{k=1}^{\infty} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)$$
Now it's looking like a telescoping series. This is more easily seen by displaying a few terms:
$$S = \left(\dfrac{2}{1} - \dfrac{4}{5}\right) + \left(\dfrac{4}{5} - \dfrac{8}{19}\right) + \left(\dfrac{8}{19} - \dfrac{16}{65}\right) + ...$$
Most of these fractions cancel each other out.
$$S = \lim_{K \rightarrow \infty} \sum_{k=1}^{K} \left(\dfrac{2^k}{3^k - 2^k} - \dfrac{2^{k + 1}}{3^{k + 1} - 2^{k + 1}}\right)= \lim_{K \rightarrow \infty} \left(2 - \dfrac{2^{K + 1}}{3^{K + 1} - 2^{K + 1}}\right) = 2$$
