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 $(1+x)^n=\sum {{n}\choose{k}} x^k$
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.