**Question:**

Prove that there is no term involving $x^{6}$ in the expansion of $\left(2 x^{2}-\frac{3}{x}\right)^{11}$.

**Solution:**

To prove: that there is no term involving $x^{6}$ in the expansion of $\left(2 x^{2}-\frac{3}{x}\right)^{11}$

Formula Used:

General term, $T_{r+1}$ of binomial expansion $(x+y)^{n}$

is given by,

$T_{r+1}={ }^{n} C_{r} x^{n-r} y^{r}$ where

${ }^{n} C_{r}=\frac{n !}{r !(n-r) !}$

Now, finding the general term of the expression, $\left(2 x^{2}-\frac{3}{x}\right)^{11}$, we get

$\mathrm{T}_{\mathrm{r}+1}={ }^{11} \mathrm{C}_{\mathrm{r}} \times\left(2 x^{2}\right)^{11-r} \times\left(\frac{-3}{\mathrm{x}}\right)^{\mathrm{r}}$

For finding the term which has ${ }^{x}{ }^{6}$ in it, is given by

$22-2 r-r=6$

$3 r=16$

$r=\frac{16}{3}$

Since, $r=\frac{16}{3}$is not possible as $r$ needs to be a whole number

Thus, there is no term involving $x^{6}$ in the expansion of $\left(2 x^{2}-\frac{3}{x}\right)^{11}$.