The smallest natural number n, such that the coefficient of x in the expansion of (x² + 1/x³)ⁿ is ⁿC₂₃ is:

Correct option is D. 58
Given (x² + 1/x³)^n
Tr = Σr=0n nCr x2n-2r.x-3r = Σr=0n nCr x2n - 5r
= 1 ⇒ 2n - 5r = 1
For r = 23
2n = 5(23) + 1 ⇒ 2n = 116 ⇒ n = 58
Smallest value of n is 58