Find the sum of series $(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+...+(n^2+1)n!$.

Question

Find the sum of series $(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+...+(n^2+1)n!$ I have found one method as i have shown in my answer below. But that form took me 30 mins to identify. $T_n=(n^2+1)n!$=$((n+1)(n+2)-3(n+1)+2)n!$ Hence adding all the terms and after cancellation the sum becomes $(n+2)!-2(n+1)!$ which simplies to $n(n+1)!$ Anyone has got better ideas out there, please let me know!

Answer 1

$$(n^2+1)n! = (n^2\color{blue}{+n-n}+1)n!= n(n+1)! - (n-1)n!$$

Answer 2

$\bf{My\; Solution}::$ We can write $$(r^2+1)r! = (r^2+2r+1-r)r!=(r+1)^2r!-r\cdot r!$$

So $$(r^2+1)r!=(r+1)\cdot (r+1)!-r\cdot r!.$$

So Sum is $\displaystyle = \sum_{r=1}^{n}\left[(r+1)\cdot (r+1)!-r\cdot r!\right] = (n+1)\cdot (n+1)!-1\cdot 1!$

Here We have used Telescopic Sum Formula.