<s>[INST] Partial fractioning an exponential function $\sum\limits_{n=1}^{\infty}\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}$ I saw this and immediately thought it was a telescoping series, so I tried to partial fraction it. $$\frac{6^n}{(3^{n+1}-2^{n+1})(3^n-2^n)}=\frac{A}{3^{n+1}-2^{n+1}}+\frac{B}{3^n-2^n}$$
\begin{align}
6^n &= A(3^n-2^n)+B(3^{n+1}-2^{n+1})\\
6^n &=3^n(A+3B)+2^n(-A-2B)
\end{align}
This is where I got stuck, how can the sum of these 2 be $6^n$ so I figured one way it could be true was if 
$$A+3B=\frac{2^n}{2}$$
$$-A-2B=\frac{3^n}{2}$$
Because then I will just be doing $\frac{6^n}{2}+\frac{6^n}{2}=6^n$
Solving this system gave me 
\begin{align}
A&=-\frac{3^{n+1}+2^{n+1}}{2}\\ B&=\frac{3^n+2^n}{2}
\end{align} 
As terrible as it looks, I got that it did telescope to $\frac{5}{2}$ however it actually came out to $2$ with the partial fraction of $$\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$
So 2 questions, 
Why am I wrong? and
What would be the correct approach to get this answer?
 [/INST] Is there a transcription problem? A check on your work shows
$$
\frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} =
\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}},
$$
and
$$
\sum _{n=1}^{\infty } \frac{6^n}{\left(3^{n+1}-2^{n+1}\right) \left(3^n-2^n\right)} = 2.
$$
</s>