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 $x^{-1}$.