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

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Solution

$T_{5}$ in ${[\sqrt[4]{2} + \frac{1}{\sqrt[4]{3}}]}^{n} = n C_{4} (\sqrt[4]{2})^{n - 4} \cdot {(\frac{1}{\sqrt[4]{3}})}^{4} = n C_{4} 2^{\frac{n - 4}{4}} \cdot \frac{1}{3} . . . (1)$

Total number of terms $= n + 1$

Fifth term from the end $= [(n + 1) - 5 + 1]^{t h}$ term from the beginning.

$\Rightarrow (n - 3)^{t h}$ term from the beginning.

$\Rightarrow T_{n - 3} = n C_{n - 4} {(\sqrt[4]{2})}^{n - (n - 4)} {(\frac{1}{\sqrt[4]{3}})}^{n - 4} = n C_{n - 4} \cdot 2 {(\frac{1}{3})}^{\frac{n - 4}{4}} . . . (2)$

Given $\frac{T_{5}}{T_{n - 3}} = \frac{\sqrt{6}}{1}$

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

$\Rightarrow \frac{n C_{4}}{n C_{n - 4}} (2)^{\frac{n - 4}{4} - 1} \cdot {(\frac{1}{3})}^{1 - \frac{n - 4}{4}} = \sqrt{6}$

$\Rightarrow \frac{n C_{4}}{n C_{n - 4}} (2)^{\frac{n - 8}{4}} \cdot {(\frac{1}{3})}^{\frac{8 - n}{4}} = \sqrt{6}$

$\Rightarrow \frac{n C_{4}}{n C_{n - 4}} (2)^{\frac{n - 8}{4}} \cdot (3)^{- \frac{8 - n}{4}} = \sqrt{6}$

$\Rightarrow \frac{n C_{4}}{n C_{n - 4}} (2)^{\frac{n - 8}{4}} \cdot (3)^{\frac{n - 8}{4}} = \sqrt{6}$

$\Rightarrow \frac{n C_{4}}{n C_{n - 4}} (6)^{\frac{n - 8}{4}} = \sqrt{6}$

Comparing powers of $6$

$\Rightarrow \frac{n - 8}{4} = \frac{1}{2}$

$\Rightarrow \frac{n - 8}{4} = \frac{1}{2}$

$\Rightarrow n - 8 = 2$

$∴ n = 10$

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