Question

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

• A. $\frac{13}{54}$
• B. $\frac{15}{54}$
• C. $\frac{17}{54}$
• D. $\frac{19}{54}$

Answer 1

The correct option is C $\frac{17}{54}$

Given, $(1+x+2x^3)(\frac{3x^2}{2}-\frac{1}{3}{)}^9$
$=\left[\right(\frac{3x^2}{2}-\frac{1}{3x}{)}^9]+x(\frac{3x^2}{2}-\frac{1}{3x}{)}^9+\left[2x^3\right(\frac{3x^2}{2}-\frac{1}{3x}{)}^9]$
Therefore for $(\frac{3x^2}{2}-\frac{1}{3}{)}^9$
$T_{r+1}=(-1{)}^r{}^9{C}_r{3}^{9-2r}{2}^{r-9}{x}^{18-3r}$
Therefore only terms indicated by the square bracket have independent term because the term not having a square bracket has an independent term for $r=\frac{19}{3}$ which is not possible.
Therefore, the independent terms are $T_7$ in first term and $T_8$ in last term.
Therefore sum of the coefficients of the independent terms is
$={}^9{C}_6{3}^{-3}{2}^{-3}-2{}^9{C}_7{3}^{-5}{2}^{-2}$
$=\frac{84}{27\left(8\right)}-\frac{2\left(36\right)}{243\left(4\right)}$
$=\frac{17}{54}$