Question

Find which term is independent of $x$ in the expansion of ${(\frac{3x^2}{2}-\frac{1}{3x})}^9$

Answer 1

General term in the expansion of $(a+b{)}^n$ is given by:

$t_{r+1}={}^n{C}_r(a{)}^{n-r}{b}^r$

General term in the expansion of ${(\frac{3x^2}{2}-\frac{1}{3x})}^9$ is

$t_{r+1}={}^9{C}_r{\left(\frac{3x^2}{2}\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r$

$t_{r+1}={}^9{C}_r\frac{3^{9-r}}{2^{9-r}}\frac{\left(x^{18-2r}\right)}{3^r{x}^r}(-1{)}^r$

$t_{r+1}={}^9{C}_r\frac{3^{9-2r}}{2^{9-r}}{x}^{28-3r}(-1{)}^r$

Term independent of $\frac{x}{0}$

$\Rightarrow x^{18-3r}=x$

$18-3r=0$

$18=3r$

$6=r$

$\therefore r=6$

$t_7={}^9{C}_6\frac{3^{-3}}{2^3}(-1{)}^6$

$t_7=\frac{9876!}{6!3!}\frac{1}{{3.2}^3}$

$t_7=\frac{7}{18}$

$\therefore 7^{th}$ term is independent of x and $t_7=\frac{7}{18}$.