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.
$$f(x)=(1+x)^n =\displaystyle \sum^{n}_{k=0}x^{k}\binom{n}{k} $$ 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 $$\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k} = \frac{(1+2)^{n+2}-1}{(n+1)(n+2)}-\frac{2}{n+1}= \frac{3^{n+2}-2n-5}{(n+1)(n+2)}$$
Hint: If $$ f(x)=\sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\, x^{k+2}\binom{n}{k} $$ then $$ f''(x)=\sum^{n}_{k=0} x^{k}\binom{n}{k} = (1+x)^n $$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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 $\ds{4\int_{0}^{1}\pars{1 - t}\pars{1 + 2t}^{n}\,\dd t = \int_{1}^{3}\pars{3x^{n} - x^{n + 1}}\dd x}$.
