The smallest natural number $n$ , such that the coefficient of $x$ in the expansion of…
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 $C_{23}^n$ , is
Solution
In the expansion of $(x^2 + \frac{1}{x^3})^n$ the general term is $T_{r+1} = C_{r}^{n} x^{2n-r} (\frac{1}{x^3})^r$
$= C_{r}^{n} x^{2n-2r-3r} = C_{r}^{n} x^{2n-5r}$
For coefficient of $x$, $2n-5r = 1$
$\Rightarrow r = \frac{2n-1}{5}$
So, we have the coefficient as $C_{\frac{2n-1}{5}}^{n}$
Using, the given value and $C_{r}^{n} = C_{n-r}^{n}$
$\Rightarrow C_{\frac{2n-1}{5}}^{n} = C_{23}^{n} = C_{n-23}^{n}$
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 'n' is 38.