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：
$(\frac {1}{100}-1)×(\frac {1}{99}-1)×(\frac {1}{98}-1)×... ×(\frac {1}{2}-1)$.
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}= $
$\frac{1}{50}$
；(2)根据上面的规律,计算：
$(\frac {1}{100}-1)×(\frac {1}{99}-1)×(\frac {1}{98}-1)×... ×(\frac {1}{2}-1)$.
解:原式$=-\frac{99}{100}×(-\frac{98}{99})×(-\frac{97}{98})×\cdots×(-\frac{1}{2})$
$=-\frac{1}{100}$.
$=-\frac{1}{100}$.
答案：1.(1)$\frac{1}{50}$
(2)解:原式$=-\frac{99}{100}×(-\frac{98}{99})×(-\frac{97}{98})×\cdots×(-\frac{1}{2})$
$=-\frac{1}{100}$.
(2)解:原式$=-\frac{99}{100}×(-\frac{98}{99})×(-\frac{97}{98})×\cdots×(-\frac{1}{2})$
$=-\frac{1}{100}$.
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:$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\cdots+\frac{1}{2022×2024}$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2022}-\frac{1}{2024})$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{2024})$
$=\frac{1}{2}×\frac{1012-1}{2024}$
$=\frac{1011}{4048}$.
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：$\frac {1}{n(n+1)}=$
$\frac{1}{n}-\frac{1}{n+1}$
.(2)直接写出下列各式的结果.
①$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{2022×2023}=$
$\frac{2022}{2023}$
；②$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{n(n+1)}=$
$\frac{n}{n+1}$
.(3)探究并计算：$\frac {1}{2×4}+\frac {1}{4×6}+\frac {1}{6×8}+... +\frac {1}{2022×2024}$.
解:$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\cdots+\frac{1}{2022×2024}$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2022}-\frac{1}{2024})$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{2024})$
$=\frac{1}{2}×\frac{1012-1}{2024}$
$=\frac{1011}{4048}$.
答案：2.(1)$\frac{1}{n}-\frac{1}{n+1}$
(2)①$\frac{2022}{2023}$ ②$\frac{n}{n+1}$
点拨:①$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{2022×2023}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2022}-\frac{1}{2023}$
$=1-\frac{1}{2023}$
$=\frac{2022}{2023}$.
②$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{n(n+1)}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}$
$=\frac{n+1-1}{n+1}$
$=\frac{n}{n+1}$.
(3)解:$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\cdots+\frac{1}{2022×2024}$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2022}-\frac{1}{2024})$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{2024})$
$=\frac{1}{2}×\frac{1012-1}{2024}$
$=\frac{1011}{4048}$.
(2)①$\frac{2022}{2023}$ ②$\frac{n}{n+1}$
点拨:①$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{2022×2023}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{2022}-\frac{1}{2023}$
$=1-\frac{1}{2023}$
$=\frac{2022}{2023}$.
②$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\cdots+\frac{1}{n(n+1)}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots+\frac{1}{n}-\frac{1}{n+1}$
$=1-\frac{1}{n+1}$
$=\frac{n+1-1}{n+1}$
$=\frac{n}{n+1}$.
(3)解:$\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\cdots+\frac{1}{2022×2024}$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{2022}-\frac{1}{2024})$
$=\frac{1}{2}×(\frac{1}{2}-\frac{1}{2024})$
$=\frac{1}{2}×\frac{1012-1}{2024}$
$=\frac{1011}{4048}$.