Show that the expansion of

For $\left(2 x^{2}-\frac{1}{x}\right)^{20}$

$a=2 x^{2}, b=\frac{-1}{x}$ and $n=20$

We have a formula,

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

$=\left(\begin{array}{c}20 \\ r\end{array}\right)\left(3 x^{2}\right)^{20-r}\left(\frac{-1}{x}\right)^{r}$

$=\left(\begin{array}{c}20 \\ r\end{array}\right)(3)^{20-r}\left(x^{2}\right)^{20-r}(-1)^{r}(x)^{-r}$

$=\left(\begin{array}{c}20 \\ \mathrm{r}\end{array}\right)(3)^{20-\mathrm{r}}(\mathrm{x})^{40-2 \mathrm{r}}(-1)^{\mathrm{r}}(\mathrm{x})^{-\mathrm{r}}$

$=\left(\begin{array}{c}20 \\ \mathrm{r}\end{array}\right)(3)^{20-\mathrm{r}}(\mathrm{x})^{40-2 \mathrm{r}-\mathrm{r}}(-1)^{\mathrm{r}}$

$=\left(\begin{array}{c}20 \\ \mathrm{r}\end{array}\right)(3)^{20-\mathrm{r}}(-1)^{\mathrm{r}}(\mathrm{x})^{40-3 \mathrm{r}}$

To get coefficient of $x^{9}$ we must have,

$(x)^{40-3 r}=(x)^{9}$

- $40-3 r=9$

- $3 r=31$

$\cdot r=10.3333$

As $\left(\begin{array}{c}20 \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}20 \\ 10.3333\end{array}\right)$ is not possible

Therefore, the term containing $x^{9}$ does not exist in the expansion of $\left(2 x^{2}-\frac{1}{x}\right)^{20}$