Question

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

Answer 1

In the expansion of $E={(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$, we have

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

$\Rightarrow T_{r+1}=(-1{)}^r{.}^9{C}_r.\frac{3^{(9-2r)}}{2^{(9-r)}}.x^{(18-3r)}.$

$\therefore (1+x+2x^3)\left[(a_0\times \frac{1}{x^3}+a_1\times \frac{1}{x}+a_2)\text{from E}\right]$

$=(1+x+2x^3)[\left\{\right(-1{)}^7{.}^9{C}_7.\frac{e^{-5}}{2^2}\times \frac{1}{x^3}\}+\left\{\right(-1{)}^6{.}^9{C}_6.\frac{3^{-3}}{2^3}\times x^0\}]$

$\begin{array}{c}x=-1\Rightarrow 18-3r=-1\Rightarrow r\text{is fraction}\\ 18-3r=0\Rightarrow r=6\text{and}18-3r=-3\Rightarrow r=7\end{array}$

$=(1+x+2x^3)[\frac{-1}{27x^3}+\frac{7}{18}]$

$\therefore$ required term $=(\frac{-2}{27}+\frac{7}{18})=\frac{17}{54}$