解$： (1)\frac {1}{n(n+1)}=\frac {1}{n}-\frac {1}{n+1}，$
$ \frac {1}{n(n+2)}=\frac {1}{2}(\frac {1}{n}-\frac {1}{n+2})。$
$ (2)\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\dots +\frac {1}{2025×2026}$
$ =1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+\dots +\frac {1}{2025}-\frac {1}{2026}$
$ =1-\frac {1}{2026}=\frac {2025}{2026}。$
$ \frac {1}{2×4}+\frac {1}{4×6}+\frac {1}{6×8}+\dots +\frac {1}{2024×2026}$
$ =\frac {1}{4}×(\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\dots +\frac {1}{1012×1013})$
$ =\frac {1}{4}×\frac {1012}{1013}=\frac {253}{1013}。$
$ (3)$原方程可化为$\frac {1}{2}(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+\dots +\frac {1}{2n-1}-\frac {1}{2n+1})=\frac {n+100}{2n+202}，$
$ $即$\frac {1}{2}(1-\frac {1}{2n+1})=\frac {n+100}{2n+202}，$
去分母，得$2n^2+202n=(2n+1)(n+100)，$
$ $解得$n=100。$
检验：因为$n$为正整数，原方程分母不会为零，
∴原方程的解为$n=100。$
$ (4)$原式$=\frac {1}{2}(1-\frac {1}{3}+\frac {1}{2}-\frac {1}{4}+\frac {1}{3}-\frac {1}{5}+\frac {1}{4}-\frac {1}{6}+\dots +\frac {1}{n-1}-\frac {1}{n+1}+\frac {1}{n}-\frac {1}{n+2})$
$ =\frac {1}{2}(1+\frac {1}{2}-\frac {1}{n+1}-\frac {1}{n+2})$
$ =\frac {n(3n+5)}{4(n+1)(n+2)}。$
