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

Answer: 9

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 $={\left(n-7+2\right)}^{th}$ term
from the beginning i.e. ${\left(n-5\right)}^{th}$ term.
Now, $\frac{\_{}^n{C}_6^{}\frac{1}{{\left(3^{\frac{1}{3}}\right)}^6}{\left(2^{\frac{1}{3}}\right)}^{n-6}}{\_{}^n{C}_{n-6}^{}\frac{1}{{\left(3^{\frac{1}{3}}\right)}^{n-6}\times {\left(2^{\frac{1}{3}}\right)}^6}}=\frac{1}{6}$
$\Rightarrow {\left(3^{\frac{1}{3}}\right)}^{n-12}\times {\left(2^{\frac{1}{3}}\right)}^{n-12}=6^{-1}$
$\Rightarrow {\left(6^{\frac{1}{3}}\right)}^{n-12}=6^{-1}$
$\Rightarrow {\left(6\right)}^{\frac{n-12}{3}}=6^{-1}$
$\Rightarrow \frac{n-12}{3}=-1$
$\Rightarrow n-12=-3$
$\Rightarrow n=9$