Evaluation of $\displaystyle \sum^{n}_{r=1}(r^2+1)\cdot r!$ using combinational argument
Although i have solved it without combinational argument
$\displaystyle \sum^{n}_{r=1}\bigg[(r+1)^2-2r\bigg]r!=\sum^{n}_{r=1}(r+1)(r+1)!-r(r!)-\sum^{n}_{r=1}\bigg[(r+1-1)r!\bigg]$
$\displaystyle \sum^{n}_{r=1}\bigg[(r+1)(r+1)!-r(r!)\bigg]-\sum^{n}_{r=1}\bigg[(r+1)!-r!\bigg]$
$\displaystyle (n+1)(n+1)!-1-(n+1)!+1=n(n+1)!$
But did not know using combinational argument
If anyone have an idea please explain me .Thanks