解：$(2)$原式$=\frac {1}{1×2}+\frac {1}{2×3}+...+\frac {1}{50×51}+\frac {1}{51×52}+\frac {1}{52×53}+\frac {1}{53×54}+…+\frac {1}{2021×2022}$
$-(\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+...+\frac {1}{50×51})$
$=\frac {2021}{2022}-\frac {50}{51}$
$=\frac {657}{34374}$
$(3)$原式$=\frac {1}{2}×(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+...+\frac {1}{2021}-\frac {1}{2023})$
$=\frac {1}{2}×(1-\frac {1}{2023})$
$=\frac {1}{2}×\frac {2022}{2023}$
$=\frac {1011}{2023}$
$(4)$解$：$原式$=1-\frac {1}{3}×(1-\frac {1}{4}+\frac {1}{4}-\frac {1}{7}+\frac 17-\frac {1}{10}$
$+...+ \frac {1}{31}-\frac {1}{34})$
$=1-\frac {1}{3}×(1-\frac {1}{34})$
$=1-\frac {1}{3}×\frac {33}{34}$
$=1-\frac {11}{34}$
$=\frac {23}{34}$
