Let the ratio of the fifth term from the beginning to the fifth term from the end in the b…

Question

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

Answer: 84

Accepted Answer

$$\left(2^{1/4}\right)^{n-8}\left(3^{-1/4}\right)^{-(n-8)}=6^{(n-8)/4}=6^{1/4}\quad\Rightarrow\quad n=9$$

$$T_6=\binom95\left(2^{1/4}\right)^4\left(3^{-1/4}\right)^5=\binom95\cdot2\cdot3^{-5/4}=\frac{84}{3^{1/4}}$$

$$\alpha=84$$

Solutions · 1