How to calculate this sum. I am trying to solve an excercise and i come across the following sum 
$ \sum_{k=1}^n \frac{(k+1)(k^3-2k+2)}{k(k+2)} $
I put it in Wolfram alpha and it says that it is equal to:
$\frac{n(2n^4+6n^3+2n^2+3n+11)}{6(n+1)(n+2)}$
but how can i prove this?

You may notice that
$$ \frac{(k+1)(k^3-2k+2)}{k(k+2)} = \left(\frac{1}{k}-\frac{1}{k+2}\right)+2\binom{k}{2} \tag{1}$$
hence by creative telescoping and the hockey stick identity
$$ \sum_{k=1}^{n}\frac{(k+1)(k^3-2k+2)}{k(k+2)} = 2\binom{n+1}{3}+\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}.\tag{2}$$
