Evaluate $\sum_{k=1}^{n}(k^2 \cdot (k+1)!)$ We have to evaluate the following:
$$1^2 \cdot 2! + 2^2 \cdot 3! + \cdots + n^2 \cdot (n+1)! =\sum_{k=1}^{n} k^2 \cdot (k+1)!$$
Any hints ?

Let
$$f_k = (k-1) (k+2)!$$
Then
$$f_k-f_{k-1} = k^2 (k+1)!$$
and
$$\begin{align}\sum_{k=1}^n k^2 (k+1)! &= f_1 - f_0 + f_2 - f_1 + \ldots+f_{n}-f_{n-1}\\ &= f_n-f_0\\  &= (n-1)(n+2)! + 2\end{align}$$
ADDENDUM
How did I get my $f_k$?  By assuming that such an anti-difference takes the form
$$f_k = (k+m) (k+n)!$$
This general form will produce a quadratic times some factorial upon differencing.  Here
$$f_k - f_{k-1} = [(k^2-(m+n-1) k + m (n-1) + 1] (k+n-1)!$$
We want to isolate a factor of $k^2$ in the brackets.  Then we have to solve
$$m+n-1=0$$
$$m (n-1) + 1=0$$
The solutions are $(m,n) = (-1,2)$ and $(1,0)$.  The former solution provides the form we seek.  
