Question: 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?

Answer: 
$$=\sum\_{k=0}^{n}\frac{k!}{k!(k+1)(k+2)}\binom{n}{k}
=\sum\_{k=0}^{n}\frac{k!}{(k+2)!}\cdot\frac{n!}{k!(n-k)!}
\\=\sum\_{k=0}^{n}\frac{n!}{(k+2)!(n-k)!}
\\=\frac{1}{(n+1)(n+2)}\sum\_{k=0}^{n}\frac{(n+2)!}{(k+2)!(n-k)!}
\\=\frac{1}{(n+1)(n+2)}\sum\_{k=0}^{n}\binom{n+2}{k+2}$$
then you can complete using $(1+x)^n=\sum {{n}\choose{k}} x^k$
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.