Mathematics · Binomial Theorem

JEE Main 2024 — 1 February, Shift 2 — Question 6

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 $\left(\frac{n}{m}\right)^{\frac{1}{3}}$ is :

Option A:
$\frac{4}{9}$
Option B:
$\frac{1}{9}$
Option C:
$\frac{1}{4}$
Option D:
$\frac{9}{4}$
Correct

Answer: D

Step-by-step solution

$\left(\frac{x^{\frac{1}{3}}}{3}+2 x^{\frac{-2}{3}}\right)^{18}$

$t_{7}={ }^{18} c_{6}\left(\frac{x^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{6}={ }^{18} c_{6} \frac{1}{(3)^{12}} \cdot \frac{1}{2^{6}}$

$t_{13}={ }^{18} c_{12}\left(\frac{x^{\frac{1}{3}}}{3}\right)^{6}\left(\frac{x^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} c_{12} \frac{1}{(3)^{6}} \cdot \frac{1}{2^{12}} \cdot x^{-6}$

$\mathrm{m}={ }^{18} \mathrm{c}_{6} \cdot 3^{-12} \cdot 2^{-6}: \mathrm{n}={ }^{18} \mathrm{c}_{12} \cdot 2^{-12} \cdot 3^{-6}$

$\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}=\left(\frac{2^{-12} \cdot 3^{-6}}{3^{-12} \cdot 2^{-6}}\right)^{\frac{1}{3}}=\left(\frac{3}{2}\right)^{2}=\frac{9}{4}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 1 February, Shift 2
Subject: Mathematics
Chapter: Binomial Theorem
Topic: Binomial Coefficients