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,
$$\sum_{r=2}^{100}...=\sum_{r=1}^{100}...$$
 A: 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.
A: Another approach is to use the geometric series starting from here:
$$
\sum_{r=2}^{100} x^{r-1}=\frac{1-x^{100}}{1-x}-1
$$
and apply all needed transformations multipling/dividing by x and integrating and derivating until you get this:
$$\sum_{r=2}^{100}\frac{x^r(2-2r)}{(r+1)(r+2)}=-\frac{2}{x^2} \int_0^x \left(\int_0^t w^2 \frac{\partial }{\partial w^1}\left(\frac{1-w^{100}}{1-w}-1\right) \, dw\right) \, dt$$
calculate the RHS since it's just 2 rational integrals and it can be done.
then substitute $x \to 3$ and the job is done.
Remark: if you want to see all the steps just look at the RHS from the inside out and start off from the first formula in the post.
steps.
Der by x $\to$ Mul by $x^2$ $\to$ Int from $0$ to $x$ $\to$ Int from $0$ to $x$ $\to$ Mul by $-\frac{2}{x^2}$
A: After the reduction to
$$ S=\sum_{r=2}^{100}\left(\frac{4\cdot 3^{r}}{r+1}-\frac{2\cdot 3^{r+1}}{r+2}\right) = \sum_{r=2}^{100}\left(\frac{3^{r+1}}{r+1}-\frac{3^{r+2}}{r+2}\right)+\sum_{r=2}^{100}\left(\frac{3^r}{r+1}+\frac{3^{r+1}}{r+2}\right) $$
we have
$$ S = 12-\frac{2\cdot 3^{101}}{51}+2\cdot\sum_{r=3}^{99}\frac{3^r}{r+1}=\boxed{1-\frac{2\cdot 3^{100}}{17}+2\cdot\color{red}{\sum_{r=0}^{99}\frac{3^{r}}{r+1}}} $$
but the last term do not telescope (it is a rational number with a huge numerator and a huge denominator, by $p$-adic heights). Anyway, it is clearly related with
$$ \int_{0}^{3}\frac{x^{100}-1}{x-1}\,dx $$
that is not difficult to approximate, since the main contribute to the integral clearly comes from a neighbourhood of $x=3$. I guess there is a typo in the original exercise, the real telescopic sum is
$$ \sum_{r=2}^{100}\frac{3^r(\color{red}{2r+1})}{(r+1)(r+2)}=\sum_{r=2}^{100}\left(\frac{3^{r+1}}{r+2}-\frac{3^r}{r+1}\right) .$$
