<s>[INST] Can anyone help me with this finite sum? I have to calculate the sum
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1}$
We can re-write the sum as follows
$\displaystyle\sum_{k=1}^n \displaystyle\frac{3^k-1+1}{(3^{k+1}-1)(3^k-1)}$
And then we obtain
$\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{3^{k+1}-1}+\displaystyle\sum_{k=1}^n \displaystyle\frac{1}{(3^{k+1}-1)(3^k-1)}$
But I don't know what to do with the last two sums. Can anyone help me with them. Thanks
 [/INST] Telescopic!
$$\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \frac{1}{2}\left(\frac{1}{3^k-1}-\frac{1}{3^{k+1}-1}\right) $$
hence:

$$ \sum_{k=1}^{n}\frac{3^k}{3^{2k+1}-3^k-3^{k+1}+1} = \color{red}{\frac{1}{2}\left(\frac{1}{2}-\frac{1}{3^{n+1}-1}\right)}.$$

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