Question:
Find the $4^{\text {th }}$ term in the expansion of $(x-2 y)^{12}$
Solution:
It is known that $(r+1)^{\text {th }}$ term, $\left(T_{r+1}\right)$, in the binomial expansion of $(a+b)^{n}$ is given by $\mathrm{T}_{\mathrm{r}+1}={ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}} \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$.
Thus, the $4^{\text {th }}$ term in the expansion of $(x-2 y)^{12}$ is
$\mathrm{T}_{4}=\mathrm{T}_{3+1}={ }^{12} \mathrm{C}_{3}(\mathrm{x})^{12-3}(-2 \mathrm{y})^{3}=(-1)^{3} \cdot \frac{12 !}{3 ! 9 !} \cdot \mathrm{x}^{9} \cdot(2)^{3} \cdot \mathrm{y}^{3}=-\frac{12 \cdot 11 \cdot 10}{3 \cdot 2} \cdot(2)^{3} \mathrm{x}^{9} \mathrm{y}^{3}=-1760 \mathrm{x}^{9} \mathrm{y}^{3}$