A) \[\frac{28}{81}\]
B) \[\frac{28}{243}\]
C) \[-\frac{28}{243}\]
D) \[-\frac{28}{81}\]
Correct Answer: B
Solution :
In \[{{\left( {{x}^{2}}-\frac{1}{3x} \right)}^{9}},\] \[{{T}_{r+1}}={{\,}^{9}}{{C}_{r}}{{({{x}^{2}})}^{9-r}}{{\left( -\frac{1}{3x} \right)}^{r}}\]\[={{\,}^{9}}{{C}_{r}}{{x}^{18-2r}}\frac{{{(-1)}^{r}}}{{{3}^{r}}}{{x}^{-r}}\] It is independent of x. \ \[18-3r=0\Rightarrow r=6\] \ \[{{T}_{7}}={{\,}^{9}}{{C}_{6}}{{x}^{18-12}}\frac{{{(-1)}^{6}}}{{{3}^{6}}}{{x}^{-6}}={{\,}^{9}}{{C}_{6}}\frac{{{(-1)}^{6}}}{36}=\frac{28}{243}\]
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