Find $S_{n}=\sum_{k=1}^{n}k!(k^2+1)$

Question

$$n\in\mathbb{N}^{*}; S_{n}=\sum_{k=1}^{n}k!(k^2+1)$$

I need to find $S_n$

I started like this: $S_{n}=\sum_{k=1}^{n}(k+2)!-3(k+1)!+2k!$

How to continue?I tried to give the k values but the terms don't vanish.

Answer 1

$$ \sum_{k=1}^{n}\bigg((k+2)!-3(k+1)!+2k!\bigg)=\sum_{k=1}^{n}\bigg(\bigg[(k+2)!-(k+1)!\bigg]-2\bigg[(k+1)!-k!\bigg]\bigg) $$ and telescope.

Answer 2

Next we use $$S_n=\sum_{k=1}^n\left\{[(k+2)!-(k+1)!]-2[(k+1)!-k!]\right\},$$which telescopes to $$(n+2)!-2-2((n+1)!-1)=(n+2)!-2\cdot(n+1)!=n\cdot (n+1)!$$

Answer 3

So far you arrived at $$\begin{array}c S_n=&&&&\hphantom{+}3!&+4!+\ldots+n!&+(n+1)!&+(n+2)!\\ &-3(&&\hphantom+2!&+3!&+4!+\ldots +n!&+(n+1)!&&)\\ &+2(&1!&+2!&+3!&+4!+\ldots +n!&&&)\end{array}$$ Can you simplify?