Question

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

Answer 1

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

$Byusinggeneralexpansion{\left(a+b\right)}^{n}is{T}_{r+1}{=}^n{C}_r{\left(a\right)}^{n-r}.{\left(a\right)}^n$

$Puttingn=6,a=\frac{3}{2x^2},b=-\frac{1}{3x}{T}_{r+1}{=}^6{C}_r{\left(\frac{3}{2}{x}^2\right)}^{6-r}{\left(-\frac{1}{3x}\right)}^r{=}^6{C}_r{\left(\frac{3}{2}\right)}^{6-r}{\left(x^2\right)}^{6-r}{\left(-\frac{1}{3}\right)}^r{\left(\frac{1}{x}\right)}^r{=}^6{C}_r{\left(\frac{3}{2}\right)}^{6-r}{\left(x\right)}^{2\left(6-r\right)}{\left(-\frac{1}{3}\right)}^r{\left(\frac{1}{x}\right)}^r{=}^6{C}_r{\left(\frac{3}{2}\right)}^{6-r}{\left(-\frac{1}{3}\right)}^r{\left(x\right)}^{12-2r-r}{=}^6{C}_r{\left(\frac{3}{2}\right)}^{6-r}{\left(-\frac{1}{3}\right)}^r{\left(x\right)}^{12-3r}----\left(1\right)$

$So,forindependentofx$

$x^{12-3r}=x^0$

$Oncomparingpower12-3r=012=3rr=4$

$Puttingr=4in\left(1\right)T_{4+1}{=}^6{C}_4{\left(\frac{3}{2}\right)}^{6-4}{\left(-\frac{1}{3}\right)}^{\left(4\right)}{\left(x\right)}^{12-3\left(4\right)}=\frac{6!}{4!\left(6-4\right)!}\left(\frac{1}{4}\right)\left(\frac{1}{9}\right)=\frac{6\times 5\times 4!}{4!.2!}\left(\frac{1}{4}\right)\left(\frac{1}{9}\right)=\frac{5}{12}$

$Hence,thetermwhichisindependentofx=T_5=\frac{5}{12}$