Binomial sum involving power of $2$ 
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$

Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\binom{n}{n}\frac{x^{n+1}}{n+1}$$
Could some help me to solve it , Thanks


We obtain
  \begin{align*}
\color{blue}{\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}2^{k+2}\binom{n}{k}}
&=\frac{1}{(n+1)(n+2)}\sum_{k=0}^n\binom{n+2}{k+2}2^{k+2}\tag{1}\\
&=\frac{1}{(n+1)(n+2)}\sum_{k=2}^{n+2}\binom{n+2}{k}2^k\tag{2}\\
&=\frac{1}{(n+1)(n+2)}\left(3^{n+2}-1-2(n+2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{(n+1)(n+2)}\left(3^{n+2}-2n-5\right)}
\end{align*}

Comment


*

*In (1) we apply the binomial identity $\binom{p+1}{q+1}=\frac{p+1}{q+1}\binom{p}{q}$ twice.

*In (2) we shift the index to start with $k=2$.

*In (3) we apply the binomial theorem.
