Find the two middle terms in the expansion of:

For $\left(x^{4}-\frac{1}{x^{3}}\right)^{11}$

$a=x^{4}, b=\frac{-1}{x^{3}}$ and $n=11$

As $n$ is odd, there are two middle terms i.e.

II. $\left(\frac{\mathrm{n}+1}{2}\right)^{\text {th }}$ and II. $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}$

General term $t_{r+1}$ is given by,

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

I. The first middle term is $\left(\frac{n+1}{2}\right)^{\text {th }}=\left(\frac{11+1}{2}\right)^{\text {th }}=\left(\frac{12}{2}\right)^{\text {th }}=(6)^{\text {th }}$

Therefore, for the $6^{\text {th }}$ middle term, $r=5$

Therefore, the first middle term is

$\mathrm{t}_{6}=\mathrm{t}_{5+1}$

$=\left(\begin{array}{c}11 \\ 5\end{array}\right)\left(x^{4}\right)^{11-5}\left(\frac{-1}{x^{3}}\right)^{5}$

$=\left(\begin{array}{c}11 \\ 5\end{array}\right)\left(x^{4}\right)^{6}(-1)^{5}\left(\frac{1}{x^{3}}\right)^{5}$

$=\left(\begin{array}{c}11 \\ 5\end{array}\right)(x)^{24}(-1) \frac{1}{x^{15}}$

$=\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \cdot(x)^{9}(-1)$

$=-462 \cdot x^{9}$

II. The second middle term is $\left(\frac{\mathrm{n}+3}{2}\right)^{\text {th }}=\left(\frac{11+3}{2}\right)^{\text {th }}=\left(\frac{14}{2}\right)^{\text {th }}=(7)^{\text {th }}$

Therefore, for the $7^{\text {th }}$ middle term, $\mathrm{r}=6$

Therefore, the second middle term is

$\mathrm{t}_{7}=\mathrm{t}_{6+1}$

$=\left(\begin{array}{c}11 \\ 6\end{array}\right)\left(x^{4}\right)^{11-6}\left(\frac{-1}{x^{3}}\right)^{6}$

$\left.=\left(\begin{array}{c}11 \\ 5\end{array}\right)\left(\mathrm{x}^{4}\right)^{5}(-1)^{6}\left(\frac{1}{\mathrm{x}^{3}}\right)^{6} \ldots \ldots \ldots\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)=\left(\begin{array}{c}\mathrm{n} \\ \mathrm{n}-\mathrm{r}\end{array}\right)\right]$

$=\left(\begin{array}{c}11 \\ 5\end{array}\right)(\mathrm{x})^{20}(1) \frac{1}{\mathrm{x}^{18}}$

$=\frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} \cdot(\mathrm{x})^{2}$

$=462 \cdot x^{2}$