A) \[\frac{5}{4}\]
B) \[\frac{4}{5}\]
C) \[27\]
D) \[182\]
Correct Answer: D
Solution :
\[{{T}_{r+1}}{{=}^{18}}{{C}_{r}}{{({{x}^{1/3}})}^{18-r}}{{\left( \frac{1}{2{{x}^{1/3}}} \right)}^{r}}\] \[=\,{{\,}^{18}}{{C}_{r}}{{\left( \frac{1}{2} \right)}^{r}}{{x}^{\frac{18-2r}{3}}}\] For coefficient of \[{{x}^{-2}},\,\,\frac{18-2r}{3}=-2\] \[\Rightarrow \] r = 12 For coefficient of \[{{x}^{-4}}\], \[\frac{18-2r}{3}=-4\] \[\Rightarrow \] \[r=15\frac{m}{m}=\frac{^{18}{{C}_{12}}{{\left( \frac{1}{2} \right)}^{12}}}{^{18}{{C}_{15}}{{\left( \frac{1}{2} \right)}^{15}}}\] \[\frac{^{18}{{C}_{6}}{{(2)}^{3}}}{^{18}{{C}_{3}}}=182\]
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