解：$(1)\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+···+\frac {1}{2021×2022}$
$=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+···+\frac {1}{2021}-\frac {1}{2022} $
$=1-\frac {1}{2022}$
$=\frac {2021}{2022} $
$(2)\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+···+\frac {1}{2021×2023} $
$=\frac {1}{2}×(1-\frac {1}{3})+\frac {1}{2}×(\frac {1}{3}-\frac {1}{5})+\frac {1}{2}×(\frac {1}{5}-\frac {1}{7})+$
$···+\frac {1}{2}×(\frac {1}{2021}- \frac {1}{2023}) $
$=\frac {1}{2}×(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+···+\frac {1}{2021}-\frac {1}{2023})$
$=\frac {1}{2}×(1-\frac {1}{2023})$
$=\frac {1011}{2023}$
$(3)$∵$\frac {1}{1+2}=\frac {1}{3}$，$\frac {1}{3}×\frac {1}{2}=\frac {1}{2×3}=\frac {1}{2}-\frac {1}{3}$；
$ \frac {1}{1+2+3}=\frac {1}{6}$，$\frac {1}{6}×\frac {1}{2}=\frac {1}{12}=\frac {1}{3×4}=\frac {1}{3}-\frac {1}{4}$；
$\frac {1}{1+2+3+4}=\frac {1}{10}$，$\frac {1}{10}×\frac {1}{2}=\frac {1}{20}=\frac {1}{4×5}=\frac {1}{4}-\frac {1}{5}$；
$\frac 1{1+2+3+4+···+2021+2022}=\frac 1{2023×1011}$
∴$\frac {1}{2023×1011}×\frac {1}{2}=\frac {1}{2022×2023}=\frac {1}{2022}-\frac {1}{2023}$
∴原式$=2×(\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+···+\frac {1}{2022×2023}) $
$=2×(\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+\frac {1}{4}-\frac {1}{5}+···+\frac {1}{2022}-\frac {1}{2023})$
$=2×(\frac {1}{2}-\frac {1}{2023})$
$=2×\frac {2021}{4046}$
$=\frac {2021}{2023}$
