Is there a closed formula for $\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}\binom{n}{k}$? I was asked to find a closed formula for the sum
$$\sum_{k=0}^{n}\frac{1}{(k+1)(k+2)}\binom{n}{k}$$
could anyone give me an advice on how to get started?
 A: $\newcommand{\+}{^{\dagger}}
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$$
\mbox{Lets consider}\quad\fermi\pars{x}\equiv
\sum_{k = 0}^{n}{x^{k + 2} \over \pars{k + 1}\pars{k + 2}}\,{n \choose k}\,,\quad
\fermi\pars{0} =0\,,\quad\fermi\pars{1} = {\large ?}
$$

$$
\fermi'\pars{x}=
\sum_{k = 0}^{n}{x^{k + 1} \over k + 1}\,{n \choose k}\,,\qquad
\fermi'\pars{0} = 0
$$

$$
\fermi''\pars{x}=
\sum_{k = 0}^{n}x^{k}{n \choose k}=\pars{1 + x}^{n}\qquad\imp\qquad
\fermi'\pars{x}={\pars{1 + x}^{n + 1} - 1\over n + 1}
$$

$$
\imp\qquad\fermi\pars{x}=
{1 \over n + 1}\,{\pars{1 + x}^{n + 2} - 1 \over n + 2} - {x \over n + 1}
$$

$$
\fermi\pars{1}
=\color{#66f}{\large\sum_{k = 0}^{n}{1 \over \pars{k + 1}\pars{k + 2}}\,
{n \choose k}
={2^{n + 2} - n - 3 \over \pars{n + 1}\pars{n + 2}}}
$$
A: Start with $(1+x)^n=\sum {{n}\choose{k}} x^k$ . If you integrate this twice wrt x you will get something close to what you are after. 
A: Hint
$$\sum_{k=0}^{n}\binom{n}{k}x^k = (1+x)^n$$ Integrate twice both rhs and lhs with respect to $x$ and when finished, plug $x=1$ in your result.
A: Sure - you know that $(1 + x)^n = \sum_{k=0} ^n \binom{n}{k} x^k$ from which it follows that 
$$
\frac{(1+x)^{n+1}}{n+1}
= \int (1 + x)^n \, \mathrm{d}x 
= \int \sum_{k=0} ^n \binom{n}{k} x^k \, \mathrm{d}x
= \sum_{k=0} ^n \binom{n}{k} \int x^k \, \mathrm{d}x
= \sum_{k=0} ^n \binom{n}{k} \frac{x^{k+1}}{k+1} .
$$
Based on what I've shown, you can iterate on this method, and allow $x$ to become a certain number, which should give you the closed form you are seeking.
A: $$=\sum_{k=0}^{n}\frac{k!}{k!(k+1)(k+2)}\binom{n}{k}
=\sum_{k=0}^{n}\frac{k!}{(k+2)!}\cdot\frac{n!}{k!(n-k)!}
\\=\sum_{k=0}^{n}\frac{n!}{(k+2)!(n-k)!}
\\=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\frac{(n+2)!}{(k+2)!(n-k)!}
\\=\frac{1}{(n+1)(n+2)}\sum_{k=0}^{n}\binom{n+2}{k+2}$$
then you can complete using $(1+x)^n=\sum {{n}\choose{k}} x^k$
