Question

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

• A. $\frac{6}{15}$
• B. $\frac{7}{18}$
• C. $\frac{7}{8}$
• D. $\frac{4}{9}$

Answer 1

Answer: (B)

$\frac{7}{18}$

We know that the general term in expansion of $(a+b{)}^n$ is given by ${}^n{C}_r{a}^{n-r}{b}^r$ where r ranges from $0$ to $9$.

Here $a=\frac{3}{2}{x}^2,b=-\frac{1}{3x}\&n=9$

For any term to be independent of $x$, coefficient of $x$ in $a^{n+1-r}{b}^{r-1}$ should be zero.
$\Rightarrow \left(x^2{)}^{9-r}\right(x^{-1}{)}^r=x^0$

$\Rightarrow 18-2r-r=0$

$\Rightarrow r=6$

Therefore, the term independent of $x$ is ${}^9{C}_6{\left(\frac{3}{2}\right)}^3{(-\frac{3}{3})}^6=\frac{7}{18}$.