Question

In the binomial expansion of $\left(\sqrt[3]{2} + \frac{1}{\sqrt[3]{3}}\right)^{n},$ the ratio of the $7^{th}$ term from the beginning to the $7^{th}$ term from the end is $1:6$ ; then find $n.$

Answer 1

We have, $\left(2^{\frac{1}{3}} + \frac{1}{3^{1 / 3}}\right)^{n}$
$7^{th}$ term from the beginning
$=\_{}^{n}C_{6}^{}\frac{1}{\left(3^{\frac{1}{3}}\right)^{6}}\left(2^{\frac{1}{3}}\right)^{n - 6}$
$7^{th}$ term from the end $=\left(n - 7 + 2\right)^{t h}$ term
from the beginning i.e. $\left(n - 5\right)^{t h}$ term.
Now, $\frac{\_{}^{n}C_{6}^{} \frac{1}{\left(3^{\frac{1}{3}}\right)^{6}} \left(2^{\frac{1}{3}}\right)^{n - 6}}{\_{}^{n}C_{n - 6}^{} \frac{1}{\left(3^{\frac{1}{3}}\right)^{n - 6} \times \left(2^{\frac{1}{3}}\right)^{6}}}=\frac{1}{6}$
$\Rightarrow \left(3^{\frac{1}{3}}\right)^{n - 12}\times \left(2^{\frac{1}{3}}\right)^{n - 12}=6^{- 1}$
$\Rightarrow \left(6^{\frac{1}{3}}\right)^{n - 12}=6^{- 1}$
$\Rightarrow \left(6\right)^{\frac{n - 12}{3}}=6^{- 1}$
$\Rightarrow \frac{n - 12}{3}=-1$
$\Rightarrow n-12=-3$
$\Rightarrow n=9$