The absolute term (that is, term independent of x) in the expansion of 32×2−13x9is:

# The absolute term (that is, term independent of $x$) in the expansion of ${\left(\frac{3}{2}{x}^{2}-\frac{1}{3x}\right)}^{9}$is:

1. A

$\frac{1}{9}$

2. B

$\frac{2}{17}$

3. C

$\frac{7}{18}$

4. D

$\frac{11}{18}$

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### Solution:

$\begin{array}{r}{T}_{r+1}{=}^{9}{C}_{r}{\left(\frac{3}{2}{x}^{2}\right)}^{9-r}{\left(-\frac{1}{3x}\right)}^{r}\\ {=}^{9}{C}_{r}{\left(\frac{3}{2}\right)}^{9-r}{\left(-\frac{1}{3}\right)}^{r}{x}^{18-3r}\end{array}$
For term independent of $x$, we set $18-3r=0$ or $r=6.$ Thus
term independent of $x$ in the expansion of ${\left(\frac{3}{2}{x}^{2}-\frac{1}{3x}\right)}^{9}$is

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