JEE Main 2022 (Online) 29th July Evening ShiftQuestion 1
If $\sum_{k=1}^{10}K^2\left({}^{10}C_K\right)^2=22000L$, then $L$ is equal to ________.
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$221$
Solution
Concept. $k\,{}^{10}C_k=10\,{}^{9}C_{k-1}$, so the sum becomes a sum of squares of ${}^{9}C_j$. Formulas used. (1) $k\,{}^{n}C_k=n\,{}^{n-1}C_{k-1}$(2) $\sum_{j=0}^{n}\left({}^{n}C_j\right)^2={}^{2n}C_n$Step 1 - Absorb. By (1), $$\sum_{k=1}^{10}k^2\left({}^{10}C_k\right)^2=\sum_{k=1}^{10}\left(10\,{}^{9}C_{k-1}\right)^2=100\sum_{j=0}^{9}\left({}^{9}C_j\right)^2.$$ Step 2 - Use (2). $\sum_{j=0}^{9}\left({}^{9}C_j\right)^2={}^{18}C_9=48620$, so the sum is $4862000$. Step 3 - Value. $22000L=4862000$ gives $L=221$. Answer: 221