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!} $
$\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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