QJEEM24Ma34a9: JEE Main 2024 Mathematics - Class 11 JEEM24Ma34a9 Solution: Let and be the coefficients of seventh and thirteenth terms…

Question

Verified Solution • 2025–26 Session moderate

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}

. Then

\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}

is :

Accepted Answer

\begin{aligned} &amp; \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} \\\\ &amp; \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}} &amp; =\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}} \\\\ &amp; =\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}

Step-by-Step Verified Solution