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, $\frac{^{n}C_{6} \left(\sqrt[3]{2}\right)^{n-6} \left(\frac{1}{\sqrt[3]{3}}\right)^{6}}{^{n}C_{n-6} \left(\sqrt[3]{2}\right)^{6} \left(\frac{1}{\sqrt[3]{3}}\right)^{n-6}} = \frac{1}{6}$
$\Rightarrow \left(2^{1/3}\right)^{n-12}\left(3^{-1/3}\right)^{12-n} = \frac{1}{6}$
$\Rightarrow \left(2^{1/3}\cdot3^{1/3}\right)^{n-12}= 6^{-1}$; or $6^{\frac{n-12}{3} } = 6^{-1}$
$\Rightarrow \frac{n-12}{3} = -1$
$\therefore n = 12 - 3 = 9$.