Question

In the binomial expansion of ${[\sqrt[3]{32}+\frac{1}{\sqrt[3]{33}}]}^n$, the ratio of the 7th term from the beginning to the 7th term from the end is $1:6$. Find $n$.

Answer 1

Given that

$\Rightarrow \frac{T_6}{T_{n-6}}=\frac{1}{6}$

$\Rightarrow \frac{{}^n{C}_6\left(2^{1/3}{)}^{n-6}\right(3^{-1/3}{)}^6}{{}^n{C}_{n-6}\left(2^{1/3}{)}^6\right(3^{-1/3}{)}^{n-6}}=\frac{1}{6}$

On simplifying we get

$\Rightarrow \left(2^{1/3}{)}^{n-6-6}\right(3^{-1/3}{)}^{6-n+6}=\frac{1}{6}$

$\Rightarrow \left(2^{1/3}{)}^{n-12}\right(3^{-1/3}{)}^{12-n}=\frac{1}{6}$

$\Rightarrow \left(2^{1/3}{)}^{n-12}\right(3^{1/3}{)}^{n-12}=\frac{1}{6}$

$\Rightarrow (6^{1/3}{)}^{n-12}=6^{-1}$

$\Rightarrow \frac{n-12}{3}=-1$

$\Rightarrow n-12=-3$

$\Rightarrow n=9$