Prove $\binom{n+2}{k+2} = \binom{n}{k+2} + 2\binom{n}{k+1} + \binom{n}{k}$ I have to prove the following statement 
$$\binom{n+2}{k+2} = \binom{n}{k+2} + 2\binom{n}{k+1} + \binom{n}{k}$$
I know, I have to use the following fact 
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$$
but I can't seem to figure out how. Some hints would be appreciated
Thanks

I kept in mind the recursive characteristics of the Pascal triangle to do this one, together with the relation you had to consider:
$$\binom{n+2}{k+2} = {\color{red}{\binom{n+1}{k+2}}} + {\color{blue}{\binom{n+1}{k+1}}} \\
= {\color{red}{\binom{n}{k+2} + \binom{n}{k+1}}}  + {\color{blue}{\binom{n}{k+1} + \binom{n}{k}}}\\
={\color{red}{\binom{n}{k+2}}} + {\color{purple}{2\binom{n}{k+1}}} + {\color{blue}{\binom{n}{k}}}$$
(The one in the middle is purple because red plus blue equals purple)
