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JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Binomial Theorem

\sum_{k = 0}^{20} (^{20} C_{k})^{2}

is equal to

(A) $^{40} C_{21}$
(B) $^{40} C_{20}$ ✓ Correct
(C) $^{41} C_{20}$
(D) $^{40} C_{19}$

Answer: (B)

Solution

By Vandermonde's identity,

\sum_{k = 0}^{n} (k n)^{2} = (n 2 n)

. Here

n = 20

, so

\sum_{k = 0}^{20} (k 20)^{2} = (20 40) =^{40} C_{20}

.

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option

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