Question

In the expansion of ${\left(2x^2-\frac{1}{x}\right)}^{12}$ , the term independent of $x$ is

• A. $8th$
• B. $7th$
• C. $9th$
• D. $10th$

Answer 1

Answer: (C)

The general term in the expansion of
${\left(2x^2-\frac{1}{x}\right)}^{12}$ is
$T_{r+1}{=}^{12}{C}_r{(2x^2)}^{12-r}\cdot {\left(-\frac{1}{x}\right)}^r$
$={{(-1)}^r}^{12}{C}_r{2}^{12-r}\cdot x^{24-2r-r}$
$={{(-1)}^r}^{12}{C}_r\cdot 2^{12-r}\cdot x^{24-3r}$
The term independent of $x$ , if $24-3r=0$
$\Rightarrow$ $24=3r$
$\Rightarrow$ $r=8$
Now, $r+1=8+1$ $=9th$ term