Question

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