Q.Find the 5th term in the expansion of $(2x-y)^{8}$.

The 5th term corresponds to $r=4$ in the general term formula, and because $r$ is even, the $(-y)^r$ part comes out positive.

The general term of $(2x+(-y))^8$ is $T_{r+1} = \,^{8}C_{r}\,(2x)^{8-r}(-y)^{r}$.

For the 5th term, $r+1=5 \Rightarrow r=4$.

$^{8}C_{4} = \dfrac{8!}{4!\,4!} = \dfrac{8\times7\times6\times5}{4\times3\times2\times1} = 70$.

$(2x)^4 = 16x^4$, and $(-y)^4 = y^4$ (an even power of a negative makes it positive).

$T_5 = 70 \times 16x^4 \times y^4 = 1120\,x^4y^4$ …

Forgetting the negative sign on $y$ and its effect — an even power of $(-y)$ is positive, but an odd-numbered term (odd $r$) would carry a negative …