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← JEE Main 2022 · 29 Jul · Shift 2
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JEE Main 2022 · 29 Jul · Shift 2, Mathematics question 7

∑r=120(r2+1)(r!)\displaystyle\sum_{r=1}^{20}(r^2+1)(r!) is equal to

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Answer and worked solution

Official answer: B

Telescope. r2+1=(r+2)(r+1)−3(r+1)+2r^2+1=(r+2)(r+1)-3(r+1)+2, so the term is f(r+1)−f(r)f(r+1)-f(r) with f(r)=(r+1)!−2 r!f(r)=(r+1)!-2\,r!.

Sum f(21)−f(1)=22!−2(21!)−0f(21)-f(1)=22!-2(21!)-0.

(B)

Topics:Special series sumsPermutations
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