Question

Find the sum of ${\sum }_{r=1}^{10}(r^2+1)(r!)$

Answer 1

${\sum }_{r=1}^{10}\left(r^2+1\right)\left(r!\right)$

$=2+5\times 2+10\times 6+17\times 24+26\times 120+37\times 720+50\times 5040+65\times 40320+82\times 362880+101\times 3628800$

$=399168000$