Question

If the ratio of the 7th term from the beginning to the 7th term from the end in the expansion of
${(2^{\frac{1}{3}}+\frac{1}{3^{\frac{1}{3}}})}^n$ is $\frac{1}{6}$, find $n$.

Answer 1

7th term from the beginning will be
$\left({}^n{C}_6{2}^{\frac{n-6}{3}}{.3}^{-2}\right)$
The 7th term from the end will be
${}^n{C}_{n-6}{2}^{\frac{6}{3}}{.3}^{\frac{6-n}{3}}$
$\therefore \frac{\left({}^n{C}_6{2}^{\frac{n-6}{3}}{.3}^{-2}\right)}{{}^n{C}_{n-6}{2}^{\frac{6}{3}}{.3}^{\frac{6-n}{3}}}=\frac{1}{6}$

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

$\Rightarrow {6.2}^{\frac{n-6}{3}}{.3}^{-2}=2^{\frac{6}{3}}{.3}^{\frac{6-n}{3}}$

$\Rightarrow {6.2}^{\frac{n}{3}}\frac{1}{36}=\frac{36}{3^{\frac{n}{3}}}$

$\Rightarrow 6^{\frac{n}{3}}=6^3$
Comparing the powers gives us
$\frac{n}{3}=3$

$\Rightarrow n=9$