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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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
$\Rightarrow r = 6$
$\Rightarrow T_{6 + 1} =^{9} C_{6} (2 x^{2})^{9 - 6} (\frac{1}{x})^{6}$
$\Rightarrow T_{7} =^{9} C_{6} 2^{3}$
$= 8^{9} C_{6}$

Suggest Corrections

Similar questions

x

{(x^{2} + \frac{1}{x})}^{9}

{(2 x^{2} - \frac{3}{x^{3}})}^{25}

x

{(\frac{3}{2} x^{2} - \frac{1}{3 x})}^{6}

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

x

{(x^{2} - \frac{1}{x})}^{9}

Join BYJU'S Learning Program