Let m and n be the coefficients of seventh and thirteenth terms respectively in the expansion of ${(\frac{1}{3} x^{\frac{1}{3}} + \frac{1}{2 x^{\frac{2}{3}}})}^{18}$ . Then ${(\frac{n}{m})}^{\frac{1}{3}}$ is:

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$\frac{9}{4}$

$\frac{1}{9}$

$\frac{4}{9}$

$\frac{1}{4}$

answer is D.

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Detailed Solution

${({\frac{x}{3}}^{\frac{1}{3}} + \frac{x^{- \frac{2}{3}}}{18})}^{18} t_{7} =^{18} c_{6} {(\frac{x^{\frac{1}{3}}}{3})}^{12} {(\frac{x^{- \frac{2}{3}}}{2})}^{6} =^{18} c_{6} \frac{1}{{(3)}^{12}} . \frac{1}{2^{6}} t_{13} =^{18} c_{12} {(\frac{x^{\frac{1}{3}}}{3})}^{6} {(\frac{x^{- \frac{2}{3}}}{2})}^{12} =^{18} c_{12} \frac{1}{{(3)}^{6}} . \frac{1}{2^{12}} . x^{- 6} m =^{18} c_{6} {.3}^{- 12} {.2}^{- 6} : n =^{18} c_{12} {.2}^{- 12} {.3}^{- 6} {(\frac{n}{m})}^{\frac{1}{3}} = {(\frac{2^{- 12} {.3}^{- 6}}{3^{- 12} {.2}^{- 6}})}^{\frac{1}{3}} = {(\frac{3}{2})}^{2} = \frac{9}{4}$