JEE Main 2019 (Online) 10th April Evening ShiftQuestion 1
The smallest natural number $n$, such that the coefficient of $x$ in the expansion of $\left(x^{2}+\frac{1}{x^{3}}\right)^{n}$ is ${}^{n}C_{23}$, is
Binomial Theorem · Coefficient of a Given Power / Term Independent of x
Solution
Concept. ${}^{n}C_r={}^{n}C_{23}$ means $r=23$ or $r=n-23$; try both. Formulas used. (1) $T_{r+1}={}^{n}C_r\,a^{n-r}b^r$ and ${}^{n}C_r={}^{n}C_{n-r}$Step 1 - General term. $T_{r+1}={}^{n}C_r\,x^{2(n-r)-3r}={}^{n}C_r\,x^{2n-5r}$. For $x^1$: $2n-5r=1$. Step 2 - Case $r=23$. $2n=116$, so $n=58$. Step 3 - Case $r=n-23$. $2n-5n+115=1$, so $n=38$. Step 4 - Smallest. $n=38$. Answer: (B)