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

$ \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$