Question

The sum ${\sum }_{r=1}^{20}(r^2+1)\times (r!)$ is equal to :-

• A. $20\times (21!)$
• B. $\left(11\right)!$
• C. $10\times (11!)$
• D. $11\times (11!)$

Answer 1

The correct option is A $20\times (21!)$

Given,

${\sum }_{r=1}^{20}(r^2+1)(r!)$

$\sum (r^{2}r!+r!)$

$\sum \left[\right(r-1\left)\right(r+1)r!+2r!]$

$\sum \left[\right(r-1\left)\right(r+1)!+2r!]$

$=[1.3!+2.4!+....+(n-1\left)\right(n+1)!]+2[1!+2!+...+n!]$

$=2.1!+2.2!+3.3!+......+n.n!+(n-1)(n+1)!$

$=1+(n-1)(n+1)!+{\sum }_{r=1}^{20}r.r!$

$=1+(n-1)(n+1)!+\sum \left[\right(r+1)r!-r!$

$=1+(n-1)(n+1)!+\left[\right(2!-1!)+(3!-2!)+....+((n+1)!-n!)$

$=1+(n-1)(n+1)!+(n+1)!-1$

$=(n+1)!n$

Putting $n=20$

we get

$=(20+1)!\times 20$

$=20\times 21!$