# Show that the expansion

Question:

Show that the expansion of $\left(x^{2}+\frac{1}{x}\right)^{12}$ does not contain any term involving $x^{-1}$.

Solution:

Suppose $x^{-1}$ occurs at the $(r+1)$ th term in the given expression.

Then,

$T_{r+1}={ }^{12} C_{r}\left(x^{2}\right)^{12-r}\left(\frac{1}{x}\right)^{r}$

$={ }^{12} C_{r} x^{24-2 r-r}$

For this term to contain $x^{-1}$, we must have

$24-3 r=-1$

$\Rightarrow 3 r=25$

$\Rightarrow r=\frac{25}{3}$

It is not possible, as $r$ is not an integer.

Hence, the expansion of $\left(x^{2}+\frac{1}{x}\right)^{12}$ does not contain any term involving $x^{-1}$.