Question: How to solve this summation?

> 
> Evaluate
>  $$
> \sum\_{k=1}^n \frac{12^k}{(4^k - 3^k)(4^{k+1} - 3^{k+1})}
> $$
> 
> 
> 

[![my attempt](https://i.stack.imgur.com/3z4Kn.jpg)](https://i.stack.imgur.com/3z4Kn.jpg)

Answer: 
You were close. You should write $$12^k=[(4^{k+1}-3^{k+1})-4(4^k-3^k)]\cdot[(4^{k+1}-3^{k+1})-3(4^k-3^k)]$$
As expected the series telescopes and therefore can be readily summed. 

From a partial fraction decomposition we have
$$\frac{1}{(4^k - 3^k)(4^{k + 1} - 3^{k + 1})} = \frac{3^{-k}}{4^k - 3^k} - \frac{4 \cdot 3^{-k}}{4^{k + 1} - 3^{k + 1}}.$$
Thus the sum becomes
$$S = \sum\_{k = 1}^n \left (\frac{4^k}{4^k - 3^k} - \frac{4^{k + 1}}{4^{k + 1} - 3^{k + 1}} \right ) = \frac{4^1}{4^1 - 3^1} - \frac{4^{n + 1}}{4^{n + 1} - 3^{n + 1}} = 4 - \frac{4^{n + 1}}{4^{n + 1} - 3^{n + 1}}.$$

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**Update**

The partial fraction decomposition can be done as follows. Let $a = 4^k$ and $b = 3^k$. Then
\begin{align\*}
\frac{12^k}{(4^k - 3^k)(4^{k + 1} - 3^{k + 1})} &= \frac{4^k \cdot 3^k}{(4^k - 3^k)(4 \cdot 4^{k} - 3 \cdot 3^{k})}\\
&= \frac{ab}{(a - b)(4a - 3b)}\\
&= \frac{a}{a - b} - \frac{4a}{4a - 3b}\\
&= \frac{4^k}{4^k - 3^k} - \frac{4^{k + 1}}{4^{k + 1} - 3^{k + 1}}.
\end{align\*}