Find $\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$ The question of course seems to be summation of a telescopic series, but despite many attempts am not able to reduce it ...

$$\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)}$$ is equal to  ??

I've tried converting $2$'s to $(3-1)$ to reduce the expression, but that too didnt work out.
Edit:
I meant that the question seems to be one of 'telescopic method', but its not necessary to do it that way.
For confirmation or as a hint the answer is:

 $$\frac{3}{2}-\frac{3^{101}}{101(102)}$$

Other might be useful information is that in this case,
$$\sum_{r=2}^{100}...=\sum_{r=1}^{100}...$$

With fraction decomposition we get $\frac{(2-2r)}{(r+1)(r+2)}=\frac{4}{r+1}-\frac{6}{r+2}$ (as mentioned by claude-leibovici), so
\begin{align}
\sum_{r=2}^{100}\frac{3^r(2-2r)}{(r+1)(r+2)} &= \sum_{r=2}^{100} 3^r \cdot (\frac{4}{r+1}-\frac{6}{r+2})\\
&= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^r \cdot 6}{r+2} \\
&= \sum_{r=2}^{100} \frac{3^r \cdot 4}{r+1}-\frac{3^{r+1} \cdot 2}{r+2} \\
\end{align}
I am sure frome here you se the telescopic series.
