Question

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

Answer 1

Given expansion is $(1+x+2x^3){(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$
Now, $T_{r+1}$ of ${(\frac{3}{2}{x}^2-\frac{1}{3x})}^9$
$T_{r+1}{=}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{(-\frac{1}{3x})}^r$
$\Rightarrow T_{r+1}=(-1{)}^r{}^9{C}_r\frac{3^{9-2r}}{2^{9-r}}{x}^{18-3r}$
Now, we will find the term independent of $x$ in the given expansion
$(1+x+2x^3)(-1{)}^r{}^9{C}_r\frac{3^{9-2r}}{2^{9-r}}{x}^{18-3r}$
$T_{r+1}=(-1{)}^r{}^9{C}_r\frac{3^{9-2r}}{2^{9-r}}{x}^{18-3r}+(-1{)}^r{}^9{C}_r\frac{3^{9-2r}}{2^{9-r}}{x}^{19-3r}+(-1{)}^r{}^9{C}_r\frac{3^{9-2r}}{2^{8-r}}{x}^{21-3r}$ .....(1)
So, the independent term in the given expansion is sum of three terms.
So first term in (1) will be independent of $x$ if
$18-3r=0$
$r=6$
So, this term will be
$t_7{=}^9{C}_6\frac{1}{6^3}$
$t_7{=}^9{C}_3\frac{1}{6^3}$
Now, second term in (1) will be independent of $x$ if
$19-3r=0$
$r=\frac{19}{3}$
which is not possible as $r$ must be a whole number.
Now, third term in (1) will be independent of $x$ if
$21-3r=0$
$r=7$
$t_8=(-1{)}^7{}^9{C}_7\frac{1}{3^5{2}^2}$
$t_8={-}^9{C}_2\frac{1}{3^5{2}^2}$
So the term independent of x in the given expansion is
${}^9{C}_3{\frac{1}{6}}^3{-}^9{C}_2\frac{1}{3^5{2}^2}$