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JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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∑r=120(r2+1)(r!)\sum_{r=1}^{20}\left(r^2+1\right)(r !) is equal to
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Correct option: B
22!−2(21!)22 !-2(21 !)
Option B (22!−2(21!)22 !-2(21 !)) is the correct answer because the summation telescopes to this result after simplification.

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