The sum _ r = 1 ^ 10 ( r^2 + 1 ) ( r! ) is equal to - Vidyadip

MCQ

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

A
$11 \times \left( {11!} \right)$
✓
$10 \times \left( {11!} \right)$
C
$\left( {11!} \right)$
D
$101 \times \left( {10!} \right)$

✓

Answer

Correct option: B.

$10 \times \left( {11!} \right)$

b
$\sum\limits_{R - 1}^{10} {({r^2} + 1)r!} $

${T_1} = ({r^2} + 1 + r - r)r! = ({r^2} + r)r! - (r - r)r!$

${T_1} = rr! + r - (r - 1)r!$

${T_1} = 1\,2! - 0$

${T_2} = 2\,3! - 1\,2!$

${T_3} = 3\,4! - 2\,3!$

${T_{10}} = 10\,11! - 9\,10!$

$\sum\limits_{R - 1}^{10} {({r^2} + 1)} \,r! = 10\,11!$

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