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The coefficient of the term independent of $x$ in the expansion of

$(1+x+2x^3)\bigg(\large\frac{3}{2}$$x^2-\large\frac{1}{3x}$$\bigg)^9$ is

$\begin{array}{1 1} \frac{1}{3} \\ \frac{19}{54} \\ \frac{17}{54} \\ \frac{1}{4} \end{array} $

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General term in the expansion of $(\large\frac{3}{2}$$x^2-\large\frac{1}{3x}$$)^9$ is
$^9C_r.(\large\frac{3}{2}$$x^2)^{9-r}.(-1)^r.(\large\frac{1}{3x})^r$
$=(-1)^r.^9C_r.\large\frac{3^{9-2r}}{2^{9-r}}$$.x^{18-3r}$
For constant term $r= 6$
$\therefore$ Constant term=$^9C_6.\large\frac{1}{216}=\large\frac{7}{18}$
For coeff. of $x^{-1}$ term $18-3r=-1$ which is not possible.
For coeff. of $x^{-3}$ term $18-3r=-3$$\Rightarrow\:r=7$
$\therefore$ Coeff of $x^{-3} =-^9C_7.\large\frac{1}{9\times 27\times 4}=-\large\frac{1}{27}$
The term independent of $x$ in the expansion of
$(1+x+2x^3).\bigg(\large\frac{3}{2}$$x^2-\large\frac{1}{3x}$$\bigg)^9$ is
=$\bigg[(1).($constant term$)\bigg]+\bigg[(1).($coeff.of $x^{-1})\bigg]+\bigg[(2)($Coeff. of $x^{-3})\bigg]$
$=\large\frac{7}{18}-\frac{2}{27}=\frac{17}{54}$
answered Oct 14, 2013 by rvidyagovindarajan_1
edited Aug 8, 2014 by sharmaaparna1
 

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