The smallest natural number n,
Question:

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 :

1. (1) 38

2. (2) 58

3. (3) 23

4. (4) 35

Correct Option: 1,

Solution:

$\left(x^{2}+\frac{1}{x^{2}}\right)^{n}$

General term $T_{r+1}={ }^{n} C_{r}\left(x^{2}\right)^{n-r}\left(\frac{1}{x^{3}}\right)^{r}={ }^{n} C_{r} \cdot x^{2 n-}$

$5 r$

To find coefficient of $x, 2 n-5 r=1$

Given ${ }^{n} C_{r}={ }^{n} C_{23} \Rightarrow r=23$ or $n-r=23$

$\therefore n=58$ or $n=38$

Minimum value is $n=38$