Show that the expansion of

Solution:

For $\left(x^{2}+\frac{1}{x}\right)^{12}$

$\mathrm{a}=\mathrm{x}^{2}, \mathrm{~b}=\frac{1}{\mathrm{x}}$ and $\mathrm{n}=12$

We have a formula,

$\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$=\left(\begin{array}{c}12 \\ r\end{array}\right)\left(x^{2}\right)^{12-r}\left(\frac{1}{x}\right)^{r}$

$=\left(\begin{array}{c}12 \\ r\end{array}\right)(x)^{24-2 r}(x)^{-r}$

$=\left(\begin{array}{c}12 \\ r\end{array}\right)(x)^{24-2 r-r}$

$=\left(\begin{array}{c}12 \\ r\end{array}\right)(x)^{24-3 r}$

To get coefficient of $x^{-1}$ we must have,

$(x)^{24-3 r}=(x)^{-1}$

- $24-3 r=-1$

- $3 r=25$

- $r=8.3333$

As $\left(\begin{array}{c}20 \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}20 \\ 8.3333\end{array}\right)$ is not possible

Therefore, the term containing $x^{-1}$ does not exist in the expansion of $\left(x^{2}+\frac{1}{x}\right)^{12}$