∑ r = 1 20 r 2 + 1 r ! is equal to

Given,

$\sum _{x=1}^{20}\left(r^2+1\right)r!$

Now on rearranging we get,

$=\sum _{x=1}^{20}\left((r+1{)}^2-2r\right)r!$

$=\sum _{x=1}^{20}\left(\left(r+1\right)\left(r+1\right)!-r·r!\right)-\sum _{r=1}^{20}r·r!$

$=\sum _{x=1}^{20}\left(\left(r+1\right)\left(r+1\right)!-r.r!\right)-\sum _{r=1}^{20}\left(\left(r+1\right)!-r!\right)$

$=\left(21.\overline{)21}-1\right)-\left(\overline{)21}-1\right)$

$=21.\overline{)21}-\overline{)21}$

$=20.21!=\left(22-2\right)21!=22!-2.21!$