 # Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of Question:

Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of $\left(\sqrt{2}+\frac{1}{\sqrt{3}}\right)^{n}$ is $\sqrt{6}: 1$

Solution:

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

Fifth term from the beginning $={ }^{n} \mathrm{C}_{4} \mathrm{a}^{n-4} \mathrm{~b}^{4}$

Fifth term from the end $={ }^{n} \mathrm{C}_{n-4} \mathrm{a}^{4} \mathrm{~b}^{n-4}$

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

${ }^{n} C_{4}(\sqrt{2})^{n-4}\left(\frac{1}{\sqrt{3}}\right)^{4}={ }^{n} C_{4} \frac{(\sqrt{2})^{n}}{(\sqrt{2})^{4}} \cdot \frac{1}{3}={ }^{n} C_{4} \frac{(\sqrt{2})^{n}}{2} \cdot \frac{1}{3}=\frac{n !}{6.4 !(n-4) !}(\sqrt{2})^{n}$ (1)

${ }^{n} C_{n-4}(\sqrt{2})^{4}\left(\frac{1}{\sqrt{3}}\right)^{n-4}={ }^{n-C} C_{n-4} \cdot 2 \cdot \frac{(\sqrt{3})^{4}}{(\sqrt{3})^{n}}={ }^{n} C_{n-4} \cdot 2 \cdot \frac{3}{(\sqrt{3})^{0}}=\frac{6 n !}{(n-4) ! 4 !} \cdot \frac{1}{(\sqrt{3})^{n}}$ (2)

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

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

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

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

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

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

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

Thus, the value of $n$ is 10 .