Question

Sum of the series ${\sum }_{r=1}^n(r^2+1)r!$ is ______

• A. $(n+1)!$
• B. $(n+2)!-1$
• C. $n(n+1)!$
• D. none of these

Answer 1

The correct option is C $n(n+1)!$

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

$\Rightarrow {\sum }_{r=1}^n(r^2+1+2r-2r)r!$

$\Rightarrow {\sum }_{r=1}^n({(r+1)}^2.r!-2r.r!)$

$\Rightarrow {\sum }_{r=1}^n({(r+1)}^{}.(r+1)!-2r.r!)$

$\Rightarrow {\sum }_{r=1}^n\left[\right(r+2-1\left)\right(r+1)!-2((r+1-1)r!\left)\right]$

$\Rightarrow {\sum }_{r=1}^n\left[\right(r+2\left)\right(r+1)!-(r+1)!-2((r+1)r!-r!\left)\right]$

$\Rightarrow {\sum }_{r=1}^n\left[\right(r+2)!-(r+1)!-2.(r+1)!+2.r!)]$

$T_1=3!-2!-2.2!+2.1!$

$T_2=4!-3!-2.3!+2.2!$

$T_3=5!-4!-2.4!+2.3!$

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$T_n=(n+2)!-(n+1)!-2.(n+1)!+2.n!$

$-------------------$

$Sum=S_n=(n+2)!-2!-2(n+1)!+2.1!$

$S_n=(n+2)!-2(n+1)!-2+2$

$S_n=(n+1)![n+2-2]$

$S_n=n.(n+1)!$

Correct answer is C