A) \[5/27\]
B) \[7/18\]
C) \[8/27\]
D) \[1/24\]
Correct Answer: B
Solution :
Key Idea: The general term of \[{{(a+x)}^{n}}\] is \[{{T}_{r+1}}{{=}^{n}}{{C}_{r}}{{a}^{n-r}}\cdot {{x}^{r}}\] The general term of\[{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\]is \[{{T}_{r+1}}{{=}^{9}}{{C}_{r}}{{\left( \frac{3}{2}{{x}^{2}} \right)}^{9-r}}{{\left( -\frac{1}{3x} \right)}^{r}}\] \[{{=}^{9}}{{C}_{r}}{{\left( \frac{3}{2} \right)}^{9-r}}{{\left( -\frac{1}{3} \right)}^{r}}{{x}^{18-3r}}\] The term is independent of\[x\] \[\therefore \] \[18-3r=0\Rightarrow r=6\] \[\therefore \] \[{{T}_{7}}{{=}^{9}}{{C}_{6}}{{\left( \frac{3}{2} \right)}^{9-6}}{{\left( -\frac{1}{3} \right)}^{6}}\] \[=\frac{9\times 8\times 7}{3\times 2\times 1}{{\left( \frac{3}{2} \right)}^{3}}{{\left( \frac{1}{3} \right)}^{6}}=\frac{7}{18}\]You need to login to perform this action.
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