Prove that $\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}$ $$\sum_{k=0}^n \binom{n}{k} k^2 = n(n+1) 2^{n-2}$$ 

If you want to use induction, prove in the following order (using Pascal's formula) that:


*

*$\displaystyle\sum_{k=0}^n \binom{n}{k}= 2^{n}$,

*$\displaystyle\sum_{k=0}^n k\binom{n}{k}= n2^{n-1}$,

*$\displaystyle\sum_{k=0}^n k^2\binom{n}{k}= n(n+1)2^{n-2}$.


For a complete proof see this.
