Question

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

• A. $\frac{7}{18}$
• B. $\frac{5}{18}$
• C. $\frac{11}{18}$
• D. $\frac{13}{18}$

Answer 1

The correct option is A $\frac{7}{18}$

We know, $T_{r+1}={}^n{C}_r{a}^{n-r}{b}^r$
Applying to the above question, we get
$T_{r+1}=(-1){}^r{}^9{C}_r.{\frac{3x^2}{2}}^{9-r}{\left(3x\right)}^{-r}$
$=(-1){}^r{}^9{C}_r{2}^{r-9}{3}^{9-2r}{x}^{18-3r}$ ...(i)
For term independent of $x$
$18-3r=0$
$\therefore r=6$
Substituting in equation (i), we get
$T_7={}^9{C}_6{2}^{-3}{3}^{-3}$ $=\frac{\mathrm{9.8.7}}{3!6^3}$ $=\frac{14}{36}$ $=\frac{7}{18}$