A) \[^{9}{{C}_{3}}.\frac{1}{{{6}^{3}}}\]
B) \[^{9}{{C}_{3}}{{\left( \frac{3}{2} \right)}^{3}}\]
C) \[^{9}{{C}_{3}}\]
D) None of these
Correct Answer: A
Solution :
In the expansion of\[{{\left( \frac{3{{x}^{2}}}{2}+\frac{1}{3x} \right)}^{9}}\], the general term is \[{{T}_{r+1}}={{\,}^{9}}{{C}_{r}}.{{\left( \frac{3{{x}^{2}}}{2} \right)}^{9-r}}{{\left( -\frac{1}{3x} \right)}^{r}}\]\[={{\,}^{9}}{{C}_{r}}{{\left( \frac{3}{2} \right)}^{9-r}}{{\left( -\frac{1}{3} \right)}^{r}}{{x}^{18-3r}}\] For the term independent of x, 18 - 3r = 0 Þ r = 6 This gives the independent term \[{{T}_{6+1}}={{\,}^{9}}{{C}_{6}}{{\left( \frac{3}{2} \right)}^{9-6}}{{\left( -\frac{1}{3} \right)}^{6}}={{\,}^{9}}{{C}_{3}}.\frac{1}{{{6}^{3}}}\]
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