Question

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

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

Answer 1

Answer: (C)

Coefficient of the term independent of $x$ is the coefficient of $x^0$ in ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9+x{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9+2x^3{\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$
Now, in ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9,$ $T_{r+1}{=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(-\frac{1}{3x}\right)}^r{=}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(-\frac{1}{3}\right)}^r{x}^{18-3r}$
$\left(i\right)$ Coefficient of $x^0\Rightarrow 18-3r=0\Rightarrow r=6$
Coefficient of $x^0$ is ${}^9{C}_6{\left(\frac{3}{2}\right)}^3{\left(-\frac{1}{3}\right)}^6$
$\left(ii\right)$ Coefficient of $x^{-1}\Rightarrow 18-3r=-1\Rightarrow r=\frac{19}{3}$ (not possible)
$\left(iii\right)$ Coefficient of $x^{-3}\Rightarrow 18-3r=-3\Rightarrow r=7$
Coefficient of $x^{-3}$ is ${}^9{C}_7{\left(\frac{3}{2}\right)}^2{\left(-\frac{1}{3}\right)}^7$
Hence, the required coefficient is $1\times {}^9{C}_6{\left(\frac{3}{2}\right)}^3{\left(-\frac{1}{3}\right)}^6+1\times 0+2\times {}^9{C}_7{\left(\frac{3}{2}\right)}^2{\left(-\frac{1}{3}\right)}^7$
$=\frac{9\times 8\times 7}{3\times 2}\times \left(\frac{+1}{8}\right)\left(\frac{1}{273}\right)+Y\times \frac{9\times 82}{2}\left(\frac{1}{4}\right)\left(-\frac{1}{24327}\right)$
$=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}$