Question

If the coefficient of x in binomial expansion of expression ${(x^2+\frac{1}{x^3})}^n$ is ${}^n{C}_{23}$. Then the minimum value of n is?

• A. $28$
• B. $48$
• C. $58$
• D. $38$

Answer 1

The correct option is D $38$

${(x^2+\frac{1}{x^3})}^n$
$T_{r+1}={}^n{C}_r\cdot (x^2{)}^{n-r}{\left(\frac{1}{x^3}\right)}^r$
$={}^n{C}_r\cdot x^{2n-2r-3r}={}^n{C}_r\cdot x^{2n-5r}$
For coefficient of $2n-5r=1$
$r=\frac{2n-1}{5}$
Coefficient of x is $={}^n{C}_{\frac{2n-1}{5}}$ or ${}^n{C}_{n-\frac{2n-1}{5}}$ (i.e. ${}^n{C}_{\frac{3n+1}{5}}$)
$\frac{2n-1}{5}=23$ $\Rightarrow 2n=116\Rightarrow n=58$
or $\frac{3n+1}{5}=23\Rightarrow 3n+1=115\Rightarrow n=38$
Minimum value of n is $38$.