Question

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

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

Answer 1

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

Given, $S=\sum _{r=1}^{10}(r^2+1)r!$

Let $T_r=(r^2+r)\times r!$

$\Rightarrow T_r=\left[r\right(r+1)-(r-1\left)\right]\times r!$

$\Rightarrow T_r=r(r+1)!-(r-1)r!$

$\therefore T_1=1(2!)-0!,T_2=2(3!)-1\left(2\right)!,T_3=3(4!)-2(3!),.......,T_{10}=10(11!)-9(10!)$

Adding all values from $T_1$ to $T_{10}$, we get

$S=10\times (11!)$

Hence, B is the right answer.