Mathematics - Binomial Theorem Question with Solution | TestHub

In the expansion of ${\left(x^2+\frac{1}{x^3}\right)}^n$ the general term is $T_{r+1}={}_{ }{}^n{C}_r{\left(x^2\right)}^{n-r}{\left(\frac{1}{x^3}\right)}^r$

$={}_{ }{}^n{C}_r{x}^{2n-2r-3r}={}_{ }{}^n{C}_r{x}^{2n-5r}$

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

$\Rightarrow r=\frac{2n-1}{5}$

So, we have the coefficient as ${}_{ }{}^n{C}_{\frac{2n-1}{5}}$

Using, the given value and ${}^{n}C_r={}^{n}C_{n-r}$

$\Rightarrow {}_{ }{}^n{C}_{\frac{2n-1}{5}}={}_{ }{}^n{C}_{23}={}_{ }{}^n{C}_{n-23}$

If $\frac{2n-1}{5}=23\Rightarrow n=58$ and if $\frac{2n-1}{5}=n-23\Rightarrow n=38$

Thus, the minimum value of $\text{'}n\text{'}$ is $38.$