Mathematics - Binomial Theorem Question with Solution | TestHub

The $r^{th}$ term of  ${\left(2x^{\frac{1}{5}}-\frac{1}{x^{\frac{1}{5}}}\right)}^{15}$

$\Rightarrow T_{r+1}={}^{15}C_r{\left(2x^{\frac{1}{5}}\right)}^{15-r}{\left(\frac{-1}{x^{\frac{1}{5}}}\right)}^r$

For term having $x^{-1}$ 

$\frac{15-r}{5}-\frac{r}{5}=-1\Rightarrow r=10$ $\Rightarrow m={}^{15}C_{10}·2^5·{\left(-1\right)}^{10}$

For term having $x^{-3}$

$\frac{15-2r}{5}=-3$ $\Rightarrow r=15$ $\Rightarrow n={}^{15}C_{15}·2^0·{\left(-1\right)}^{15}$

$\therefore m{n}^2={}^{15}C_{10}·2^5={}^{15}C_5·2^5$

On comparing with $m·n^2={}^{15}C_r·2^r$, we get $r=5$