Question

If the term independent of $x$ in the expansion of ${\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$ is $k$, then $18k$ is equal to :

• A. 9
• B. 11
• C. 5
• D. 7

Answer 1

Answer: (D)

$T_{r+1}={}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(-\frac{1}{3x}\right)}^r$
$T_{r+1}={}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(-\frac{1}{3}\right)}^r{x}^{18-3r}$
For independent of $x$
$18-3r=0,r=6$
$\therefore T_7={}^9{C}_6{\left(\frac{3}{2}\right)}^3{\left(-\frac{1}{3}\right)}^6=\frac{21}{54}=k$
$\therefore 18k=\frac{21}{54}\times 18=7$