解:$(2)$根据$a_{1}=-3$,$a_{2}=\frac {1}{4}$,$a_{3}=\frac {4}{3}$,$a_{4}=-3$,..可知,
$a_{1}$,$a_{2}$,$a_{3}$,...这列数每三个数为一个循环.
因为$2024÷3=674……2$,$a_{1}+a_{2}+a_{3}=-3+\frac {1}{4}+\frac {4}{3}=-\frac {17}{12}$,
所以$-\frac {17}{12}×674=-\frac {5729}{6}.$
所以$a_{1}+a_{2}+a_{3}+a_{4}+...+a_{2021}+a_{2022}+a_{2023}+a_{2024}=-\frac {5729}{6}+(-3)+\frac {1}{4}=-\frac {11491}{12}$