Question

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

• A. $\frac{17}{54}$
• B. $\frac{17}{52}$
• C. $\frac{19}{54}$
• D. $\frac{19}{52}$

Answer 1

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

For independent term in expansion
$(1+x+2x^3){(\frac{3x^2}{2}-\frac{1}{3x})}^9$
it can be written as $\underset{\left(1\right)}{{(\frac{3x^2}{2}-\frac{1}{3x})}^9}+\underset{\left(2\right)}{x{(\frac{3x^2}{2}-\frac{1}{3x})}^9}+\underset{\left(3\right)}{2x^3{(\frac{3x^2}{2}-\frac{1}{3x})}^9}$

Now, solving each for independent term of $x$
$(r+1{)}^{\text{th}}$ term for ${(\frac{3x^2}{2}-\frac{1}{3x})}^9$
$T_{r+1}={}^9{C}_r{\left(\frac{3}{2}{x}^2\right)}^{9-r}{\left(\frac{-1}{3x}\right)}^r={}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(\frac{-1}{3}\right)}^r{x}^{18-3r}$
Now, for $\left(1\right)$ to be independent fom $x$
$18-3r=0\Rightarrow r=6$
So, ${}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(\frac{-1}{3}\right)}^r=\frac{7}{18}$

Similarly, for $\left(2\right)$ to be independent from $x$
$18-3r=-1\Rightarrow r=\frac{19}{3}$ (not possible)

For $\left(3\right)$ to be independent from $x$
$18-3r=-3\Rightarrow r=7$
$\Rightarrow {}^9{C}_r{\left(\frac{3}{2}\right)}^{9-r}{\left(\frac{-1}{3}\right)}^r=-\frac{1}{27}$

Hence, the required term will be
$\frac{7}{18}-2\cdot \frac{1}{27}=\frac{17}{54}$