In (3√2+13√3)n if the ratio of 7th term from the beginning to th

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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 =

Options:

A . 7

B . 8

C . 9

D . None of these

Answer: Option C
:
C

$\frac{1}{6} = \frac{^{n} C_{6} {(2^{\frac{1}{3}})}^{n - 6} {(3^{- \frac{1}{3}})}^{6}}{^{n} C_{n - 6} {(2^{\frac{1}{3}})}^{6} {(3^{- \frac{1}{3}})}^{n - 6}}$
$6^{- 1} = 6^{\frac{1}{3} (n - 12)}$
$n - 12 = - 3$
$n = 9$

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