A) \[\frac{1}{3}\]
B) \[\frac{19}{54}\]
C) \[\frac{17}{54}\]
D) \[\frac{1}{4}\]
Correct Answer: C
Solution :
The \[{{r}^{th}}\] term in the expansion of \[{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\] is \[{{T}_{r+1}}{{=}^{9}}{{C}_{r}}{{\left( \frac{3}{2}{{x}^{2}} \right)}^{9-r}}{{\left( -\frac{1}{3x} \right)}^{r}}\] \[{{=}^{9}}{{C}_{r}}{{\left( \frac{3}{2} \right)}^{9-r}}{{\left( -\frac{1}{3} \right)}^{r}}{{x}^{18-3r}}\] ?.(i) The coefficient of the term independent of x in the expansion of \[(1+x+2{{x}^{3}})\,\,{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\] ?..(ii) = Sum of the coefficient of the terms \[{{x}^{0}},\] \[{{x}^{-1}}\] and \[{{x}^{-3}}\] in \[{{\left( \frac{3}{2}{{x}^{2}}-\frac{1}{3x} \right)}^{9}}\] For \[{{x}^{0}}\] in (i) above, \[18-3r=0\Rightarrow r=6.\]. for \[{{x}^{-1}}\]in (i) above, there exists no value ofr and hence no such term exists. For \[{{x}^{-3}}\]in (i), \[18-3r=-3\Rightarrow r=7\] \[\therefore \] for term independent of x, in (ii) the coefficient \[=1{{\times }^{9}}{{C}_{6}}{{(-1)}^{6}}{{\left( \frac{3}{2} \right)}^{9-6}}{{\left( \frac{1}{3} \right)}^{6}}+2{{\times }^{9}}{{C}_{7}}{{(-1)}^{7}}{{\left( \frac{3}{2} \right)}^{9-7}}{{\left( \frac{1}{3} \right)}^{7}}\]\[=\frac{98.7}{1.2.3}.\frac{{{3}^{3}}}{{{2}^{3}}}.\frac{1}{{{3}^{6}}}+2\frac{9.8}{1.2}(-1)\frac{{{3}^{2}}}{{{2}^{2}}}.\frac{1}{{{3}^{7}}}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}.\]
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