Binomial Theorem Question - None

Question

In the expansion $\left(\sqrt{2} \sqrt[6]{3}+\frac{1}{\sqrt[3]{3}}\right)^{n}$, if the ratio of 7^th term from the beginning to the 7^th term from the end is $\frac{1}{6}$, then n =

Accepted Answer

7^th term is $\left[\sqrt[6]{3} \sqrt{2}+\frac{1}{\sqrt[3]{3}}\right]^{n}$ = ⁿC₆ $\left(2^{\frac{1}{2}} 3^{\frac{1}{6}}\right)^{n-6}\left[\frac{1}{3^{\frac{1}{ 3}}}\right]^{6}$
7^th term from the end in $\left[\sqrt[5]{3} \sqrt{2}+\frac{1}{\sqrt[3]{3}}\right]^{n}$
= ⁿC₆ $\left(\frac{1}{3^{\frac{1}{3}}}\right)^{n-6}\left(3^{\frac{1}{ 6}} 2^{\frac{1}{ 2}}\right)^{6}$
$\frac{{ }^{n} \mathrm{C}_{6}\left[2^{\frac{1}{ 2}} 3^{\frac{1}{ 6}}\right]^{n-6}\left[\frac{1}{3^{\frac{1}{ 3}}}\right]^{6}}{{ }^{n} \mathrm{C}_{6}\left[\frac{1}{3^{\frac{1}{ 3}}}\right]^{n-6}\left[2^{\frac{1}{ 2}} 3^{\frac{1}{ 6}}\right]^{6}}=\frac{1}{6}$
$\Leftrightarrow \frac{2^{\frac{n-6}{2}} 3^{\frac{n-6}{6}}}{3^{\frac{6-n}{3}} 2^{3} \cdot 3^{1}} \cdot \frac{1}{3^{2}}=\frac{1}{6}$
$\Leftrightarrow \frac{2^{\frac{n-6}{2}}}{3^{\frac{6-n}{3}}} \cdot \frac{1}{\left.3^{-\left(\frac{n-6}{6}\right.}\right)} \cdot \frac{1}{2^{3} 3^{3}}=\frac{1}{6}$
$\Rightarrow 2^{\frac{n-6}{2}} .3^{\frac{n-6}{2}}$ = 2² $\times$ 3²
$\Rightarrow \frac{n-6}{2}$ = 2 $\Rightarrow$ n = 10

Question:

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