જો $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{n}$ ના વિસ્તરણના શરૂઆતથી પાંચમા પદ અને છેલ્લે થી પાંચમા પદનો ગુણોત્તર $\sqrt{6}: 1$ હોય, તો $n$ શોધો.

In the expansion, ${(a + b)^n} = {\,^n}{C_0}{a^n} + {\,^n}{C_1}{a^{n - 1}}b + {\,^n}{C_2}{a^{n - 2}}{b^2} +  \ldots  + {\,^n}{C_{n - 1}}a{b^{n - 1}} + {\,^n}{C_n}{b^n}$

Fifth term from the beginning $ = {\,^n}{C_4}{a^{n - 4}}{b^4}$

Fifth term from the end $ = {\,^n}{C_4}{a^4}{b^{n - 4}}$

Therefore, it is evident that in the expansion of $\left(\sqrt[4]{2}+\frac{1}{\sqrt[4]{3}}\right)^{ n }$ are fifth term from the beginning is ${\,^n}{C_4}{(\sqrt[4]{2})^{n - 4}}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^4}$ and the fifth term from the end is ${\,^n}{C_{n - 4}}{(\sqrt[4]{2})^4}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}}$

${\,^n}{C_4}{(\sqrt[4]{2})^{n - 4}}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^4} = {\,^n}{C_4}\frac{{{{(\sqrt[4]{2})}^n}}}{{{{(\sqrt[4]{2})}^4}}} \cdot \frac{1}{3} = \frac{{n!}}{{6.4!(n - 4)!}}{(\sqrt[4]{2})^n}$           ............$(1)$

${\,^n}{C_{n - 4}}{(\sqrt[4]{2})^4}{\left( {\frac{1}{{\sqrt[4]{3}}}} \right)^{n - 4}} = {\,^n}{C_{n - 4}}\frac{{{{(\sqrt[4]{3})}^4}}}{{{{(\sqrt[4]{3})}^n}}} = {\,^n}{C_{n - 4}} \cdot 2 \cdot \frac{3}{{{{(\sqrt[4]{3})}^n}}} = \frac{{6n!}}{{(n - 4)!4!}} \cdot \frac{1}{{{{(\sqrt[4]{3})}^n}}}$         ..........$(2)$

It is given that the ratio of the fifth term from the beginning to the fifth term from the and is $\sqrt{6}: 1 .$ Therefore, from $(1)$ and $(2),$ we obtain

$\frac{ n !}{6.4 !( n -4) !}(\sqrt[4]{2})^{ n }: \frac{6 n !}{( n -4) ! 4 !} \cdot \frac{1}{(\sqrt[4]{3})^{ n }}=\sqrt{6}: 1$

$\Rightarrow \frac{(\sqrt[4]{2})^{n}}{6}: \frac{6}{(\sqrt[4]{3})^{n}}=\sqrt{6}: 1$

$\Rightarrow \frac{(\sqrt[4]{2})^{n}}{6} \times \frac{(\sqrt[4]{3})^{n}}{6}=\sqrt{6}$

$\Rightarrow(\sqrt[4]{6})^{n}=36 \sqrt{6}$

$ \Rightarrow {6^{n/4}} = {6^{5/2}}$

$\Rightarrow \frac{n}{4}=\frac{5}{2}$

$\Rightarrow n =4 \times \frac{5}{2}=10$

Thus, the value of $n$ is $10$