JEE Mains · Maths · STD 11 - 6. permutation and combination
The sum \(\sum\limits_{r = 1}^{10} {\left( {{r^2} + 1} \right)} \times \left( {r!} \right)\) is equal to
- A \(11 \times \left( {11!} \right)\)
- B \(10 \times \left( {11!} \right)\)
- C \(\left( {11!} \right)\)
- D \(101 \times \left( {10!} \right)\)
Answer & Solution
Correct Answer
(B) \(10 \times \left( {11!} \right)\)
Step-by-step Solution
Detailed explanation
\(\sum\limits_{R - 1}^{10} {({r^2} + 1)r!} \) \({T_1} = ({r^2} + 1 + r - r)r! = ({r^2} + r)r! - (r - r)r!\) \({T_1} = rr! + r - (r - 1)r!\) \({T_1} = 1\,2! - 0\) \({T_2} = 2\,3! - 1\,2!\) \({T_3} = 3\,4! - 2\,3!\) \({T_{10}} = 10\,11! - 9\,10!\)…
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