Mathematical Reflections 2013, Issue 1 - Problem J256

Problem:
Evaluate $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!.$$

Proposed by Titu Andreescu.

Solution:
We have $$\begin{array}{lcl} k^2(k+1)! & = & [(k^2+k-2)-(k-2)](k+1)!\\ & = & [(k-1)(k+2)-(k-2)](k+1)!\\ & = &(k-1)(k+2)!-(k-2)(k+1)! \end{array}$$ for all $k \in \mathbb{N}$. Therefore, $$1^2 2!+2^2 3! + \ldots + n^2(n+1)!=(n-1)(n+2)!+2.$$