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 ?

Using $k^2 = (k+3)(k+2) - 5 (k+2) + 4$ we write $$f_k = k^2 (k+1)! = (k+3)! - 5 (k+2)! + 4 (k+1)! = G_{k+1} - G_k$$
where $G_k = (k+2)! - 4(k+1)!$, therefore
$$
   \sum_{k=1}^n f_k = \sum_{k=1}^n \left(G_{k+1} - G_k\right) = G_{n+1} - G_1
$$
Since $G_1 = -2$, and $G_{n+1} = (n+3)! - 4(n+2)! = (n+2)! (n-1)$ we get
$$
    \sum_{k=1}^n k^2 (k+1)! = (n-1) (n+2)! + 2
$$
