Question

The smallest natural number $n,$ such that the cofficient of $x$ in the expansion of ${(x^2+\frac{1}{x^3})}^n$is ${}^n{C}_{23}$, is :

• A. $23$
• B. $38$
• C. $35$
• D. $58$

Answer 1

The correct option is B $38$

Given,
${(x^2+\frac{1}{x^3})}^n$, its $(r+1{)}^{th}$ term, is

$T_{r+1}={}^n{C}_r{x}^{2n-2r}\cdot x^{-3r}$
$={}^n{C}_r{x}^{2n-5r}$

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

$\therefore \text{Coeff. of}x={}^n{C}_{\left(\frac{2n-1}{5}\right)}={}^n{C}_{23}$
$\therefore \frac{2n-1}{5}=23$ or $n-\left(\frac{2n-1}{5}\right)=23$
$\Rightarrow n=58$ or $n=38$