Question

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

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

Answer 1

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

$\begin{array}{c}\sum \left(r^2+1\right)r!=\sum \left[r\left(r+1\right)-\left(r-1\right)\right]r!\\ ={\sum }_{r=1}^{10}\left[r\left(r+1\right)!-\left(r-1\right)r!\right]=\left(1\times 2!-0\times 1!\right)+\left(2\times 3!-1\times 2!\right)\\ +.................+\left(10\times 11!-9\times 10!\right)\\ =10\times 11!\end{array}$