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

Answer: 10

$T_5={}^n{C}_4(\sqrt[4]{2}{)}^{n-4}{\left(\frac{1}{\sqrt[4]{3}}\right)}^4={}^n{C}_4(2{)}^{\frac{n-4}{4}}\left(\frac{1}{3}\right)$
$5^{\text{th }}$ 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]
$\Rightarrow \frac{{}^n{C}_4(2{)}^{\frac{n-4}{4}\left(\frac{1}{3}\right)}}{{}^n{C}_{n-4}(2)\left(\frac{1}{3^{\frac{n-4}{4}}}\right)}=\sqrt{6}$
$\Rightarrow \left(2^{\frac{n-8}{4}}\right)\left(3^{\frac{n-8}{4}}\right)=\sqrt{6}$
$\Rightarrow (6{)}^{\frac{n-8}{4}}=6^{\frac{1}{2}}$
$\Rightarrow \frac{n-8}{4}=\frac{1}{2}$
$\Rightarrow n=10$