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 :
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 $n=38$
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