Question

In the expansion $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}$, if the ratio of the $5^{\text {th }}$ term from the beginning to the $5^{\text {th }}$ term from the end is $\frac{\sqrt{6}}{1}$, find $n$.

Answer 1

$T _{5}={ }^{n} C _{4}(\sqrt[4]{2})^{n-4}\left(\frac{1}{\sqrt[4]{3}}\right)^{4}={ }^{n} C _{4}(2)^{\frac{n-4}{4}}\left(\frac{1}{3}\right)$
$5^{\text {th }}$ term from the end $=(n-5+2)^{\text {th }}$ term from the beginning
$=(n-3)^{\text {th }}$ term from the beginning
$T _{n-3}={ }^{n} C _{n-4}(\sqrt[4]{2})^{n-n+4}\left(\frac{1}{\sqrt[4]{3}}\right)^{n-4}$
$={ }^{n} C _{n-4}(2)\left(\frac{1}{3^{\frac{n-4}{4}}}\right)$
$\frac{ T _{5}}{ T _{n-3}}=\frac{\sqrt{6}}{1}$ ...[Given]
$\Rightarrow \frac{{ }^{n} C_{4}(2)^{\frac{n-4}{4}\left(\frac{1}{3}\right)}}{{ }^{n} C_{n-4}(2)\left(\frac{1}{3^{\frac{n-4}{4}}}\right)}=\sqrt{6}$
$\Rightarrow\left(2^{\frac{n-8}{4}}\right)\left(3^{\frac{n-8}{4}}\right)=\sqrt{6}$
$\Rightarrow(6)^{\frac{n-8}{4}}=6^{\frac{1}{2}}$
$\Rightarrow \frac{n-8}{4}=\frac{1}{2}$
$\Rightarrow n=10$