JEE Main 2021 (Online) 27th August Morning ShiftQuestion 1
$\sum_{k=0}^{20}\left({}^{20}C_k\right)^{2}$ is equal to :
Binomial Theorem · Properties and Sums of Binomial Coefficients
Solution
Concept. The sum of squares of the coefficients of $(1+x)^n$ is the middle coefficient of $(1+x)^{2n}$. Formulas used. (1) $\sum_{k=0}^{n}\left({}^{n}C_k\right)^2={}^{2n}C_n$Step 1 - Why (1). Write $\left({}^{n}C_k\right)^2={}^{n}C_k\,{}^{n}C_{n-k}$. The sum is then the coefficient of $x^n$ in $(1+x)^n(1+x)^n=(1+x)^{2n}$. Step 2 - Apply with $n=20$. $\displaystyle\sum_{k=0}^{20}\left({}^{20}C_k\right)^2={}^{40}C_{20}$. Answer: (C)