Answer
$-40r^2s^3$
Work Step by Step
Using the $(r+1)$st term of the expansion of $(a+b)^n$, which is given by $\dfrac{n!}{(n-r)!r!}a^{n-r}b^r$, then the $
4
$th term of $
(2r-s)^5
$ is
\begin{array}{l}
\dfrac{5!}{2!3!}(2r)^{2}(-s)^{3}
\\\\=
10(4r^2)(-s^3)
\\=
-40r^2s^3
\end{array}