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

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Solution

The correct option is C
9

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

∴ $\frac{n}{3}$ - 4 = -1 ⇒ n = 9.

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