Question

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

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

Answer 1

Answer: (B)

$ \displaystyle\sum_{r=1}^{10}\left(r^{2}+1\right) .r ! $
$= \displaystyle\sum_{r=1}^{10}\left\{(r+1)^{2}-2 r\right\} r !$
$= \displaystyle\sum_{r=1}^{10}(r+1)(r+1) !-2 \displaystyle\sum_{r=1}^{10} r .r ! $
$= \displaystyle\sum_{r=1}^{10}\{(r+1)(r+1) !-r(r !)\}-\displaystyle\sum_{r=1}^{10} r .r !$
$=(11.11 !-1)-\displaystyle\sum_{r=1}^{10}((r+1) !-r !) $
$=(11.11 !-1-(11 !-1 !)$
$= 10.11 ! $