Question

If, in the expansion of ${\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)}^n$, the ratio of the seventh term from the beginning to the seventh term from the end is equal to $\frac{1}{6}$, then $n$ is equal to

• A. $3$
• B. $6$
• C. $9$
• D. None of these

Answer 1

Answer: (C)

We have, $T_7={}^n{C}_6{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^6$
$7^{th}$ term from the end
$=T_{n-7+2}=T_{n-5}={}^n{C}_{n-6}{\left(\sqrt[3]{2}\right)}^6{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}$
Given, $\frac{{}^n{C}_6{\left(\sqrt[3]{2}\right)}^{n-6}{\left(\frac{1}{\sqrt[3]{3}}\right)}^6}{{}^n{C}_{n-6}{\left(\sqrt[3]{2}\right)}^6{\left(\frac{1}{\sqrt[3]{3}}\right)}^{n-6}}=\frac{1}{6}$
$\Rightarrow {\left(2^{1/3}\right)}^{n-12}{\left(3^{-1/3}\right)}^{12-n}=\frac{1}{6}$
$\Rightarrow {\left(2^{1/3}\cdot 3^{1/3}\right)}^{n-12}=6^{-1}$; or $6^{\frac{n-12}{3}}=6^{-1}$
$\Rightarrow \frac{n-12}{3}=-1$
$\therefore n=12-3=9$.