Question

The coefficient of the term independent of x in the expansion of $(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

The general term in the expansion of ${(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$ is
$T_{(}r+1){=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{(-\frac{1}{3x})}^r{=}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}(-\frac{1}{3})x^{18-3r}$
Now, the coefficient of the term independent of x in the expansion of $(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$ = Sum of the coefficient of the terms $x^0,x^{-1}$ and $x^{-3}$ in ${(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$ .....(ii)
For $x^0$ in (i) above, $18–3r=0\Rightarrow r=6.x^{-1}$ in (i) above, there exists no value of r and hence no such term exits. For $x^{-3}$ in (i), $18–3r=–3\Rightarrow r=7$
$\therefore$ For term independent of x, in (ii) the coefficient
$=1{\times }^9{C}_6(-1{)}^6{\left(\frac{3}{2}\right)}^{9-6}{\left(\frac{1}{3}\right)}^6+2{\times }^9{C}_7(-1{)}^7{\left(\frac{3}{2}\right)}^{9-7}{\left(\frac{1}{3}\right)}^7$
$=\frac{\mathrm{9.8.7}}{\mathrm{1.2.3}}.\frac{3^3}{2^3}.\frac{1}{3^6}+2\frac{9.8}{1.2}(-1)\frac{3^2}{2^2}.\frac{1}{3^7}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}$