<s>[INST] Find the sum of series $(1^2+1)1!+(2^2+1)2!+(3^2+1)3!+...+(n^2+1)n!$. 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! 
 [/INST] $\bf{My\; Solution}::$ 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.
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