$$\sum_{r=1}^{20}(r^2+1)(r!)$$ is equal to

Given,

$$S=\sum_{r=1}^{20}(r^2+1)r!$$

Now,

$$r^2+1=(r+1)^2-2r$$

Hence,

$$S=\sum_{r=1}^{20}\left((r+1)^2-2r\right)r!$$

$$=\sum_{r=1}^{20}(r+1)^2r!-2\sum_{r=1}^{20}rr!$$

Using

$$ (r+1)^2r!=(r+1)(r+1)! $$

and

$$ rr!=(r+1)!-r! $$

we get

$$S=\sum_{r=1}^{20}(r+1)(r+1)!-2\sum_{r=1}^{20}\left((r+1)!-r!\right)$$

Now,

$$ (r+1)(r+1)!=(r+2-1)(r+1)! $$

$$=(r+2)!-(r+1)!$$

Therefore,

$$S=\sum_{r=1}^{20}\left((r+2)!-(r+1)!\right)-2\sum_{r=1}^{20}\left((r+1)!-r!\right)$$

Both sums are telescopic.

Hence,

$$S=(22!-2!)-2(21!-1!)$$

$$=22!-2-2\cdot21!+2$$

$$=22!-2\cdot21!$$

$$=21!(22-2)$$

$$=22!\ -\ 2.21!$$

Hence,

$$\boxed{22!\ -\ 2.21!}$$.