1. 观察下列各式：
①$\frac {1}{2}×\frac {2}{3}= \frac {1}{3}$；②$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}= \frac {1}{4}$；③$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×\frac {4}{5}= \frac {1}{5}$；$\cdot \cdot \cdot \cdot \cdot \cdot$
(1)猜想$\frac {1}{2}×\frac {2}{3}×\frac {3}{4}×... ×\frac {49}{50}= $；
(2)根据上面的规律,计算：
$(\frac {1}{100}-1)×(\frac {1}{99}-1)×(\frac {1}{98}-1)×... ×(\frac {1}{2}-1)$.

2. 阅读下列材料,然后回答问题.
观察下列等式：$\frac {1}{1×2}= 1-\frac {1}{2},\frac {1}{2×3}= \frac {1}{2}-\frac {1}{3},\frac {1}{3×4}= \frac {1}{3}-\frac {1}{4},... ... $
将以上三个等式相加得
$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}$
$=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}$
$=1-\frac {1}{4}$
$=\frac {3}{4}$.
(1)猜想并写出：$\frac {1}{n(n+1)}=$.
(2)直接写出下列各式的结果.
①$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{2022×2023}=$；
②$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{n(n+1)}=$.
(3)探究并计算：$\frac {1}{2×4}+\frac {1}{4×6}+\frac {1}{6×8}+... +\frac {1}{2022×2024}$.
解:$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\cdots+\frac{1}{2022×2024}$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2022}-\frac{1}{2024})$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{2024})$
$=\frac{1}{2}×\frac{1012-1}{2024}$
$=\frac{1011}{4048}$.