Q48 / 90 · JEE Main 2021 (27 Aug Shift 1) · Mathematics

$\sum_{k = 0}^{20} {({}^{20}C_{k})}^{2}$ is equal to

A
${}^{40}C_{21}$
B
${}^{41}C_{20}$
Correct answer
C
${}^{40}C_{20}$
D
${}^{40}C_{19}$

Solution

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

Sum of suffix is constant so summation will be ${}^{40}C_{20}$

Hence, the correct answer is option: 'C'.

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