Mathematics - Binomial Theorem Question with Solution | TestHub

Given expansion is${\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)}^n$

We know that general term of an expansion${\left(x+a\right)}^n$is$T_{r+1}={}^{n}C_r{x}^{n-r}{a}^r$

($5^{\mathrm{th}}$ term from beginning)

$T_{4+1}={}^{n}C_4(\sqrt[4]{2}{)}^{n-4}·{\left(\frac{1}{\sqrt[4]{3}}\right)}^4$

($5^{\mathrm{th}}$ term from end)

We know that$r$th term from the end in a binomial expansion is$T_{n-r+2}$from the beginning.

$T_5^{\text{'}}=T_{n-4+1}={}^{n}C_4{\left(\frac{1}{\sqrt[4]{3}}\right)}^{n-4}(\sqrt[4]{2}{)}^4$

It is also given that the ratio of the fifth term from the beginning to the fifth term from the end is $\sqrt{6}:1$.

Now, $\frac{T_5}{T_5^{\text{'}}}=\frac{\sqrt{6}}{1}$

$\Rightarrow 2^{\frac{n-8}{4}}·3^{\frac{n-8}{4}}=\sqrt{6}$

$\therefore (\sqrt{6}{)}^{\frac{n-8}{2}}=(\sqrt{6})$

$\therefore n=10$

$\therefore T_3={}^{10}C_2{\left(\sqrt[4]{2}\right)}^8{\left(\frac{1}{\sqrt[4]{3}}\right)}^2=\frac{45\times 4}{\sqrt{3}}$

$=60\sqrt{3}$

Hence, this is the correct option.