Question

Find the term independent of $x$ in the expansion of ${(\frac{3}{2}{x}^2-\frac{1}{3x})}^6$

Answer 1

${(\frac{3}{2}{x}^2-\frac{1}{3x})}^6$

$\Rightarrow {(r+1)}^{th}$ term $={{}^{6}C}_r{\left(\frac{3}{2}{x}^2\right)}^{6-r}{(-\frac{1}{3x})}^r$

$={{}^{6}C}_r{\left(\frac{3}{2}\right)}^{6-r}{(-\frac{1}{3})}^r\frac{x^{12-2r}}{x^r}$

So, as term is independent of $x.$

i.e, $x^{12-3r}=x^0$

$\Rightarrow 12-3r=0$

$\Rightarrow r=4$

So, term $5^{th}$

Hence, solved.