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STD 11 - 6. permutation and combination — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6. permutation and combination4 Marks
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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