Mathematics GRE

Is there a simple formula for this sum?

$$\sum_{k=0}^nk(n-k)=0\cdot n+1\cdot (n-1)+2\cdot(n-2)+\cdots+(n-1)\cdot 1+n\cdot 0$$

It looks kind of familiar .. something to prove with induction in a proof writing class.
Thanks

Distribute the k, break into difference of two sums... see what you can do.

$$\sum_{k=0}^{n}k(n-k)=n\sum_{k=0}^{n}k-\sum_{k=0}^{n}k^2=\frac{n^2(n+1)}{2}-\frac{n(n+1)(2n+1)}{6}=\frac{(n-1)n(n+1)}{6}$$