Question

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

• A. $\frac{1}{3}$
• B. $\frac{19}{54}$
• C. $\frac{17}{54}$
• D. $\frac{1}{4}$

Answer 1

The correct option is C $\frac{17}{54}$

The $(r+1)th$ term in the expansion of ${[\frac{3}{2}{x}^2-\frac{x}{3}]}^9$ is given by
$T_{r+1}{=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{(-\frac{1}{3x})}^r{=}^9{C}_r{(-1)}^r\frac{3^{9-2r}}{2^{9-r}}{x}^{18-3r}$ ...(1)
Since we are looking for the coefficient of the term independent of x in expansion of
$(1+x+{2x}^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$ ...(2)
We must get coefficient of $x^0,x^{-1}$ and $x^{-3}$ in the expansion of ${[\frac{3}{2}{x}^2-\frac{x}{3}]}^9$
For $x^0$, $r$ must be $6$ in (1);

for $x^{-1}$ there is no value of $r$,

and for $x^{-3}$, $r$ must be $7$ in (1)
Therefore the coefficient of the term independent of $x$ in (2) is
${1.}^9{C}_r{(-1)}^6.\frac{3^{9-12}}{2^{9-6}}+{2.}^9{C}_r{(-1)}^7.\frac{3^{9-14}}{2^{9-7}}$
$=\frac{\mathrm{9.8.7}}{\mathrm{1.2.3}}.\frac{3^{-3}}{2^3}+2.\frac{9.8}{1.2}(-1).\frac{3^{-5}}{2^2}=\frac{7}{18}-\frac{12}{27}=\frac{17}{54}$