Closed form of the summation- $\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}$ I have got the following summation-$$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}.$$
I have to find the closed form or the general form to find the sum of this series. I know upto summation of Telescopic Series and Some special series like those in $AP$ or $GP$.
But I have no idea on how to begin on this problem.
Thanks for any help!! 

Since you requested a more detailed answer than that of @N.S, here you go.  
We start with the series you give, 
$$\displaystyle\sum_{r=1}^{n}\frac{r^24^r}{(r+1)(r+2)}$$
We now want to simplify this into a series you can deal with; the first step will be to run partial fraction decomposition on the term 
$$\frac{r^2}{(r+1)(r+2)}=\frac{r^2+3r+2-3r-2}{(r+1)(r+2)}=\frac{(r+1)(r+2)-3r-2}{(r+1)(r+2)}$$
$$=1-\frac{3r+2}{(r+1)(r+2)} =1+\frac{1}{r+1}-\frac{4}{r+2}$$
Let me know if you need help with this partial fraction decomposition. We can now rewrite the above series as
$$\displaystyle\sum_{r=1}^{n}\left(4^r+\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)=\displaystyle\sum_{r=1}^{n}\left(4^r\right)+\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)\tag{1}$$
The first series is geometric, which you say you know how to evaluate. (It comes out to be $\frac{4}{3}(4^n - 1)$)  
To evaluate the second series in $(1)$ we will look at the expanded terms:
$$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right)$$
$$= \left(\frac{4}{2}\color{red}{-\frac{4^2}{3}}\right)+\left(\color{red}{\frac{4^2}{3}}\color{green}{-\frac{4^3}{4}}\right)+\left(\color{green}{\frac{4^3}{4}}\color{purple}{-\frac{4^4}{5}}\right)+\cdots+\left(\color{olive}{\frac{4^{n-1}}{n}}\color{fuchsia}{-\frac{4^n}{n+1}}\right)+\left(\color{fuchsia}{\frac{4^n}{n+1}}\color{navy}{-\frac{4^{n+1}}{n+2}}\right)$$
As my above, colored terms hopefully make clear almost all the terms will cancel out with their neighboring term except for the first and last terms (this is called a telescoping series) and so this series comes out to be
$$\sum_{r=1}^{n}\left(\frac{4^r}{r+1}-\frac{4^{r+1}}{r+2}\right) = \frac{4}{2}-\frac{4^{n+1}}{n+2}=2-\frac{4^{n+1}}{n+2}$$
We can now go plug this back into $(1)$ and rewrite your original series as
$$\frac{4}{3}(4^n - 1)+2-\frac{4^{n+1}}{n+2}= 4^{n+1}\left(\frac{1}{3}-\frac{1}{n+2}\right)+\frac{2}{3}$$
Note that, as $n\to\infty$, this expression becomes closer and closer to being $$4^{n+1}\left(\frac{1}{3}-0\right)+\frac{2}{3}=\frac{4^{n+1}+2}{3}$$
And so the limit towards $+\infty$ clearly diverges to $+\infty$; likewise, the limit towards $-\infty$ diverges towards $-\infty$. 
