Find `n`, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of `(root(4)2+1/(root(4)3))^n is\ sqrt(6): 1`.

Explanation

In the expansion, `(a+b)^{n}=\ ^{n} C_{0} a^{n}+\ ^{n} C_{1} a^{n-1} b+\ ^{n} C_{2} a^{a-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_{n-4} a^{4} b^{n-4}`

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

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

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[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`

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