If the ratio of the fifth term from the begin | vedclass

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Question

$\frac{{ }^{ n } C _4 2^{\frac{ n -4}{4}} \cdot\left(3^{\frac{-1}{4}}\right)^4}{{ }^{ n } C _4 3^{-\left(\frac{ n -4}{4}\right)} \cdot\left(2^{\frac{1

Vedclass · Accepted Answer

$60 \sqrt{3}$

Vedclass · Answer

$\frac{{ }^{ n } C _4 2^{\frac{ n -4}{4}} \cdot\left(3^{\frac{-1}{4}}\right)^4}{{ }^{ n } C _4 3^{-\left(\frac{ n -4}{4}\right)} \cdot\left(2^{\frac{1}{4}}\right)^4}=\frac{\sqrt{6}}{1}$

$\Rightarrow n =10$

So $T_3={ }^{10} C _2 2^{\frac{1}{4} \cdot 8} \cdot 3^{-\frac{1}{4}-2}=\frac{45.4}{\sqrt{3}}=60 \sqrt{3}$

If the ratio of the fifth term from the begining to the fifth term from the end in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^n$ is $\sqrt{6}: 1$, then the third term from the beginning is:

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