Let m and n be the coefficients of seventh and thirteenth terms respectively in the expans…

Question

Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$

Accepted Answer

$\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^6\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} \mathrm{C}_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot \mathrm{x}^{-6}\end{aligned}$

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

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

$\begin{aligned}\left(\frac{m}{n}\right)^{\frac{1}{3}} & =\left(\frac{{ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6}{{ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}}\right)^{\frac{1}{3}} \\ & =\left(\frac{\left(\frac{1}{3}\right)^6}{\left(\frac{1}{2}\right)^6}\right)^{\frac{1}{3}}=\left(\left(\frac{2}{3}\right)^6\right)^{\frac{1}{3}}=\frac{4}{9}\end{aligned}$

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

Solutions · 1