Find the term independent of x in the expansion of $(2 x^{2} + \frac{1}{x})^{9}$

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Solution

The correct option is B

$T_{r + 1} =^{9} C_{r} (2 x^{2})^{9 - r} (\frac{1}{x})^{r}$
The term independent of x is the term which has the power of x equals to zero. Consider the power of x in the general term or $T_{r + 1}$
= $X^{18 - 2 r} X^{- r}$
= $X^{18 - 3 r}$
Equating the power of x to zero,
18-3r=0
Or 3r=18
$\Rightarrow r = 6$
$\Rightarrow T_{6 + 1} =^{9} C_{6} (2 x^{2})^{9 - 6} (\frac{1}{x})^{6}$
$\Rightarrow T_{7} =^{9} C_{6} 2^{3}$
$= 8^{9} C_{6}$

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