JEE PYQ: Binomial Theorem Question 41
Question 41 - 2019 (10 Apr Shift 2)
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 ${}^nC_{23}$, is:
(1) 38 (2) 58 (3) 23 (4) 35
Show Answer
Answer: (1)
Solution
$2n - 5r = 1$ with ${}^nC_r = {}^nC_{23}$. If $n - r = 23$: $n = 38$. Smallest is 38.