The value of  $\sum _{k=1}^{\infty }\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}$  can be equal to;

Let $3^k=x$ 
$\therefore \frac{2^k=y}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\frac{xy}{(x-y)(3x-2y)}$ 
$=\frac{y(3x-2y)-2y(x-y)}{(x-y)(3x-2y)}=\frac{y}{x-y}-\frac{2y}{3x-2y}=\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}}$ 
$\therefore \sum _{k=1}^{\infty }\frac{6^k}{(3^k-2^k)(3^{k+1}-2^{k+1})}=\sum _{k=1}^{\infty }\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}}$ 
$T_1=\frac{2}{3-2}-\frac{2^2}{3^2-2^2}, T_2=\frac{2^2}{3^2-2^2}-\frac{2^3}{3^3-2^3}$ 
$T_k=\frac{2^k}{3^k-2^k}-\frac{2^{k+1}}{3^{k+1}-2^{k+1}}$ 
$\sum _{k=1}^{\infty }{T}_k=\underset{k\to \infty }{\lim }(2-\frac{{2.2}^k}{{3.3}^k-2·2^k})=2$