Infinite summation with multiple terms in denominator and variables in exponents

Question

I need to find $\sum\limits_{n=1}^{\infty} \frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\cdot 6^n}$, but I don't know how to do infinite summation with multiple terms in the denomiator and variables in exponents. Can anybody give me a hint? Thanks :)

Answer 1

Hint

Let $x=2^n$ and $y=3^n$ to make $$\frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\times 6^n}=\frac{x y}{2 x^2-5 x y+3 y^2}=\frac{x y}{(2 x-3 y) (x-y)}$$ Now, as @person commented, using partial fractions $$\frac{x y}{(2 x-3 y) (x-y)}=\frac{x}{y-x}-\frac{2 x}{3 y-2 x}=\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$ which seems to telescope.