Find a formula for $\sum\limits_{r=1}^{n} (r^2+1)(r!)$

Question

The sum $$\sum\limits_{r=1}^{n} (r^2+1)(r!)$$ is equal to:

1. $(n+1)!$
2. $(n+2)!-1$
3. $n\cdot(n+1)!$
4. $n\cdot(n+2)!$

My work. I tried to solve this problem by converting $(r^2+1)$ in squares then applying the property but i was unable to get the solution, please help?

Answer 1

Since there is a factorial, it's better to rewrite $r^2+1$ as suggested above by Winther's comment: $$r^2 + 1 = (r+2)(r+1) - 3(r+1) + 2,$$ then $$\begin{align}(r^2+1)r!=(r+2)(r+1)r! - 3(r+1)r! + 2r!&=(r+2)! - 3(r+1)! + 2r!\\ &=((r+2)! - (r+1)!) - 2((r+1)!-r!). \end{align}$$ Now split the sum in two parts and try to simplify. What do you obtain?