Question

The sum $\sum _{r=1}^{10}(r^2+1)\times (r!)$ is equal to:

• A. $10\times (11!)$
• B. $11\times (11!)$
• C. $(11!)$
• D. $101\times (10!)$

Answer 1

The correct option is A $10\times (11!)$

$\sum _{r=1}^{10}(r^2+1)\times (r!)$
$=\sum _{r=1}^{10}\left(\right(r+1{)}^2-2r)\times (r!)$
$=\sum _{r=1}^{10}(r+1)(r+1)!-2\sum _{r=1}^{10}r\times (r!)$
$=\sum _{r=1}^{10}\left\{\right(r+1\left)\right(r+1)!-r\times (r!\left)\right\}-\sum _{r=1}^{10}r\times (r!)$
$=(11\times (11!)-1)-\sum _{r=1}^{10}r\times (r!)$
$=(11\times (11!)-1)-\sum _{r=1}^{10}(r+1-1)\times (r!)$
$=(11\times (11!)-1)-\left(\sum _{r=1}^{10}\right(r+1)!-\sum _{r=1}^{10}r!)$
$=(11\times (11!)-1)-(11!-1!)$
$=11\times (11!)-11!$
$=10\times 11!$