Mathematics - Binomial Theorem Question with Solution | TestHub

Given expansion is ${\left(\frac{1}{3}{x}^{\frac{1}{3}}+\frac{1}{2x^{{\displaystyle \frac{2}{3}}}}\right)}^{18}$.

$\Rightarrow T_7={}^{18}C_6{\left(\frac{1}{3}{x}^{\frac{1}{3}}\right)}^{12}{\left(\frac{1}{2x^{{\displaystyle \frac{2}{3}}}}\right)}^6$

$\Rightarrow m={}^{18}C_6{\left(\frac{1}{3}\right)}^{12}{\left(\frac{1}{2}\right)}^6$

$\Rightarrow T_{13}={}^{18}C_{12}{\left(\frac{1}{3}{x}^{\frac{1}{3}}\right)}^6{\left(\frac{1}{2x^{{\displaystyle \frac{2}{3}}}}\right)}^{12}$

$\Rightarrow n={}^{18}C_{12}{\left(\frac{1}{3}\right)}^6{\left(\frac{1}{2}\right)}^{12}$

$\Rightarrow \frac{m}{n}=\frac{{}^{18}C_6{\left(\frac{1}{3}\right)}^{12}{\left(\frac{1}{2}\right)}^6}{{}^{18}C_{12}{\left(\frac{1}{3}\right)}^6{\left(\frac{1}{2}\right)}^{12}}$

$\Rightarrow \frac{m}{n}=\frac{{\left(\frac{1}{3}\right)}^6}{{\left(\frac{1}{2}\right)}^6}$

$\Rightarrow \frac{m}{n}={\left(\frac{2}{3}\right)}^6$

$\Rightarrow {\left(\frac{m}{n}\right)}^{\frac{1}{3}}=\frac{4}{9}$

$\Rightarrow {\left(\frac{n}{m}\right)}^{\frac{1}{3}}=\frac{9}{4}$