例 计算:$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{79×81}$
思路点拨 仔细研究会发现,每一个分数都可以写成$(\frac {1}{A}-\frac {1}{B})×\frac {1}{B-A}$的形式,如$\frac {1}{1×3}= (1-\frac {1}{3})×$$\frac {1}{2},\frac {1}{3×5}= (\frac {1}{3}-\frac {1}{5})×\frac {1}{2}$,提取$\frac {1}{2}$后,我们先将每一个分数转化成两个分数的差,再将能够抵消的项消去,即可得到答案。
答案详解 $\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{79×81}$
$=(1-\frac {1}{3})×\frac {1}{2}+(\frac {1}{3}-\frac {1}{5})×\frac {1}{2}+... +(\frac {1}{79}-\frac {1}{81})×\frac {1}{2}$
$=(1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+... +\frac {1}{79}-\frac {1}{81})×\frac {1}{2}$
$=(1-\frac {1}{81})×\frac {1}{2}$
$=\frac {80}{81}×\frac {1}{2}$
$=\frac {40}{81}$

$\frac{40}{81}$

1. 计算:$\frac {1}{1×3}+\frac {1}{3×5}+\frac {1}{5×7}+... +\frac {1}{99×101}$

$\frac{50}{101}$

2. 计算:$\frac {1}{1×4}+\frac {1}{4×7}+\frac {1}{7×10}+... +\frac {1}{22×25}$

$\frac{8}{25}$