How do you find the value of $\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $? How do you find the value of:
$$\sum_{r=1}^{\infty} \frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})} $$
I tried to apply partial fractions/divide numerator and denominator by $6^r$ and use the fact that $a^x \rightarrow 0$ as $x \rightarrow \infty$ for $0<a<1$, but couldn't use it because I couldn't form a telescoping series.
I can't think of anything else. Any help will be appreciated. 

Hint:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
   = \frac{2^r}{3^r-2^r} - \frac{2^{r+1}}{3^{r+1}-2^{r+1}}.$$
Edit:
$$\frac{6^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}
    = \frac{A}{3^r-2^r} + \frac{B}{3^{r+1}-2^{r+1}}
    = \frac{(3A+B)3^r - (2A+B)2^r}{(3^r-2^r)(3^{r+1} - 2^{r+1})}.$$
One obvious way to make the numerators equal is for $3A+B = 2^r$ and $2A+B=0$, which gives $A = 2^r$, $B = -2\cdot 2^r = -2^{r+1}$. (The other obvious possibility, $3A+B=0$ and $2A+B=-3^r$, gives a different decomposition which still telescopes, but where the "leftover" term at the end doesn't tend to zero).
