Sum of the following series upto n terms:$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$ Find the sum of the following series up to n terms:
$\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k$
My attempt:
First,I tried to apply binomial series,but i cannot understand how to apply it.
Then ,I tried as follows:
  $\sum_{k=1}^n \frac {k}{(k+1)(k+2)} 2^k=\sum_{k=1}^n \frac{\frac{2k+3}{2}-\frac{3}{2}}{(k+1)(k+2)}2^k=\sum_{k=1}^n\big(\frac{1}{2}-\frac{3}{2(k+1)(k+2)}\big)2^k=\sum_{k=1}^n2^{k-1}-\frac{3\cdot2^{k-1}}{(k+1)(k+2)}$
Now if we find $\sum_{k=1}^n \frac{2^{k-1}}{(k+1)(k+2)}$ then we are done.but I cannot find this.
Am I going right?If not,how should I proceed?
Thanks

It is a telescopic series. We have:
$$ \frac{k}{(k+1)(k+2)} = \frac{2}{k+2}-\frac{1}{k+1}$$
hence:
$$ \frac{k}{(k+1)(k+2)}2^k = \frac{2^{k+1}}{k+2}-\frac{2^k}{k+1} $$
and:
$$ \sum_{k=1}^{n}\frac{k}{(k+1)(k+2)}2^k = -1+\frac{2^{n+1}}{n+2}.$$
