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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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## 1 Answer

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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.

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