JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If $\left({}^{40}C_0\right)+\left({}^{41}C_1\right)+\left({}^{42}C_2\right)+\ldots+\left({}^{60}C_{20}\right)=\frac{m}{n}\,{}^{60}C_{20}$, $m$ and $n$ are coprime, then $m+n$ is equal to $\_\_\_\_\_\_$ .
Binomial Theorem · Properties and Sums of Binomial Coefficients
Answer$102$
Solution
Concept. ${}^{40+i}C_i$ for $i=0,\dots,20$ is a hockey-stick sum. Formulas used. (1) $\sum_{i=0}^{k}{}^{m+i}C_i={}^{m+k+1}C_k$Step 1 - Sum. By (1) with $m=40$, $k=20$: $$\sum_{i=0}^{20}{}^{40+i}C_i={}^{61}C_{20}.$$ Step 2 - Compare with ${}^{60}C_{20}$. $${}^{61}C_{20}=\frac{61!}{20!\,41!}=\frac{61}{41}\cdot\frac{60!}{20!\,40!}=\frac{61}{41}\,{}^{60}C_{20}.$$ Step 3 - Value. $61$ and $41$ are coprime, so $m+n=102$. Answer: 102