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← JEE Main 2022 · 28 Jun · Shift 1
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JEE Main 2022 · 28 Jun · Shift 1, Mathematics question 1

If ∑k=131(31Ck)(31Ck−1)−∑k=130(30Ck)(30Ck−1)=α(60!)(30!)(31!)\displaystyle\sum_{k=1}^{31}\left({}^{31}C_k\right)\left({}^{31}C_{k-1}\right)-\sum_{k=1}^{30}\left({}^{30}C_k\right)\left({}^{30}C_{k-1}\right)=\frac{\alpha(60!)}{(30!)(31!)}, where α∈R\alpha\in\mathbb R, then the value of 16α16\alpha is equal to

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Answer and worked solution

Official answer: A

Vandermonde. ∑knCk nCk−1=∑knCk nCn−k+1=2nCn+1\sum_k{}^nC_k\,{}^nC_{k-1}=\sum_k{}^nC_k\,{}^nC_{n-k+1}={}^{2n}C_{n+1}.

So the expression is 62C32−60C31{}^{62}C_{32}-{}^{60}C_{31}. Dividing by 60!30! 31!\dfrac{60!}{30!\,31!}: 62⋅6132−30=141116\dfrac{62\cdot61}{32}-30=\dfrac{1411}{16}.

16α=141116\alpha=1411.

(A)

Topics:Properties of binomial coefficients
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