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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$.......$
$84$
$83$
$82$
$86$
Solution
$\frac{T_{5}}{T_{n-1}}=\frac{{ }^{n} C_{4}\left(2^{1 / 4}\right)^{n-4}\left(3^{-1 / 4}\right)^{4}}{\left(2^{1 / 4}\right)^{4}\left(3^{-1 / 4}\right)^{n-4}}=\frac{\sqrt[4]{6}}{1}$
$\Rightarrow 2^{\frac{n-8}{4}} 3^{\frac{n-8}{4}}=6^{1 / 4}$
$\Rightarrow 6^{n-3}=6$
$\Rightarrow n-8=1 \Rightarrow n=9$
$T_{6}={ }^{9} C_{5}\left(2^{1 / 4}\right)^{4}\left(3^{-1 / 4}\right)^{5}=\frac{84}{\sqrt[4]{3}}$
$\therefore \alpha=84$
