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
 A: 
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.
A: $$f(x)=(1+x)^n =\displaystyle \sum^{n}_{k=0}x^{k}\binom{n}{k} $$
Thus
$$g(x)=\int f(x) dx = \frac{(1+x)^{n+1}-1}{n+1}= \displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)}\cdot x^{k+1}\binom{n}{k}$$
Where the constant of integration was chosen so that $g(0)=0$. Also,
$$\int g(x) dx= \frac{(1+x)^{n+2}-1}{(n+1)(n+2)}-\frac{x}{n+1}=\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot x^{k+2}\binom{n}{k}$$
So that $$\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k} =  \frac{(1+2)^{n+2}-1}{(n+1)(n+2)}-\frac{2}{n+1}=  \frac{3^{n+2}-2n-5}{(n+1)(n+2)}$$
A: Hint: If
$$
f(x)=\sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\, x^{k+2}\binom{n}{k}
$$
then
$$
f''(x)=\sum^{n}_{k=0} x^{k}\binom{n}{k} = (1+x)^n
$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[10px,#ffd]{\ds{%
\sum_{k=0}^{n}{1 \over \pars{k + 1}\pars{k + 2}}\, 2^{k + 2}{n \choose k}}} =
\sum_{k=0}^{n}\pars{{1 \over k + 1} - {1 \over k + 2}}\, 2^{k + 2}{n \choose k} \\[5mm] = &\
\sum_{k=0}^{n}\bracks{\int_{0}^{1}\pars{t^{k} - t^{k + 1}}\dd t}\, 2^{k + 2}{n \choose k} =
4\int_{0}^{1}\pars{1 - t}\sum_{k=0}^{n}{n \choose k}\pars{2t}^{k}\,\dd t
\\[5mm] & =
4\int_{0}^{1}\pars{1 - t}\pars{1 + 2t}^{n}\,\dd t =
\bbx{3^{n + 2} - 2n - 5 \over \pars{n + 2}\pars{n + 1}}
\end{align}

Set $\ds{x = 1 + 2t \iff t = {x - 1 \over 2}}$ such that
  $\ds{4\int_{0}^{1}\pars{1 - t}\pars{1 + 2t}^{n}\,\dd t = \int_{1}^{3}\pars{3x^{n} - x^{n + 1}}\dd x}$.

