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Mathematics Question on Sum of First n Terms of an AP

The sum ∑r=110(r2+1)×(r!)\displaystyle\sum^{10}_{r=1}(r^2 + 1) \times (r!) is equal to :

A

(11!)(11 !)

B

10×(11!)10 \times (11!)

C

101×(11!)101 \times (11!)

D

11×(11!)11 \times (11!)

Answer

10×(11!)10 \times (11!)

Explanation

Solution

∑r=110(r2+1).r!\displaystyle\sum_{r=1}^{10}\left(r^{2}+1\right) .r ! = \displaystyle\sum_{r=1}^{10}\left\\{(r+1)^{2}-2 r\right\\} r ! =∑r=110(r+1)(r+1)!−2∑r=110r.r!= \displaystyle\sum_{r=1}^{10}(r+1)(r+1) !-2 \displaystyle\sum_{r=1}^{10} r .r ! =∑r=110(r+1)(r+1)!−r(r!)−∑r=110r.r!= \displaystyle\sum_{r=1}^{10}\\{(r+1)(r+1) !-r(r !)\\}-\displaystyle\sum_{r=1}^{10} r .r ! =(11.11!−1)−∑r=110((r+1)!−r!)=(11.11 !-1)-\displaystyle\sum_{r=1}^{10}((r+1) !-r !) =(11.11!−1−(11!−1!)=(11.11 !-1-(11 !-1 !) =10.11!= 10.11 !