Question

Let the ratio of the fifth term from the beginning to the fifth term from the end in the binomial expansion of ${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^n$, in the increasing powers of $\frac{1}{\sqrt[4]{3}}$ be $\sqrt[4]{6}:1$. If the sixth term from the beginning is $\frac{\alpha }{\sqrt[4]{3}}$, then $\alpha$ is equal to

Answer 1

Answer: 84

$\frac{T_5}{T_{n-3}}=\frac{{}^n{C}_4{\left(2^{1/4}\right)}^{n-4}{\left(3^{-1/4}\right)}^4}{{}_{n-4}{\left(2^{1/4}\right)}^4{\left(3^{-1/4}\right)}^{n-4}}=\frac{\sqrt[4]{6}}{1}$
$\Rightarrow 2^{\frac{n-8}{4}}{3}^{\frac{n-8}{4}}=6^{1/4}$
$\Rightarrow 6^{n-8}=6$
$\Rightarrow n-8=1\Rightarrow n=9$
$T_6={}^9{C}_5{\left(2^{1/4}\right)}^4{\left(3^{-1/4}\right)}^5=\frac{84}{\sqrt[4]{3}}$
$\therefore \alpha =84$