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Find n if the ratio of the 5th term from the beginning to the 5th term from the end in the expansion of (4√2 + 1/4√3)ⁿis √6:1?

$F i f t h t e r m f r o m t h e e n d = (n + 1 - 5 + 1)^{t h} t e r m f r o m t h e b e g i n n i n g {=(n-3)}^{t h} t e r m f r o m t h e b e g i n n i n g t_{5} =^{n} C_{4} (\sqrt[4]{2})^{n - 4} {(\frac{1}{\sqrt[4]{3}})}^{4} =^{n} C_{4} \times 2^{\frac{n - 4}{4}} \times \frac{1}{3} t_{n - 3} =^{n} C_{n - 4} (\sqrt[4]{2})^{n - (n - 4)} {(\frac{1}{\sqrt[4]{3}})}^{n - 4} =^{n} C_{n - 4} \times 2 \times (\frac{1}{3^{\frac{n - 4}{4}}}) \frac{t_{5}}{t_{n - 3}} = \frac{\sqrt{6}}{1} \Rightarrow \frac{{}^{n}{C_{4} \times 2^{\frac{n - 4}{4} \times \frac{1}{3}}}}{{}^{n}{C_{n - 4} \times 2 \times 3^{- \frac{n - 4}{4}}}} = \frac{\sqrt{6}}{1} \Rightarrow 2^{\frac{n - 4}{4} - 1} \times 3^{\frac{n - 4}{4} - 1} = \sqrt{6} \Rightarrow 2^{\frac{n - 8}{4}} \times 3^{\frac{n - 8}{4}} = (6)^{\frac{1}{2}} \Rightarrow (2 \times 3)^{\frac{n - 8}{4}} = (6)^{\frac{1}{2}} \Rightarrow {(6)}^{\frac{n - 8}{4}} = (6)^{\frac{1}{2}} \Rightarrow \frac{n - 8}{4} = \frac{1}{2} \Rightarrow n = 10$

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