计算：$\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + … + \frac{1}{98×99} + \frac{1}{99×100}$。
思路导引：
$\frac{1}{1×2}$可以写成$1 - \frac{1}{2}$，$\frac{1}{2×3}$可以写成$\frac{1}{2} - \frac{1}{3}$，依此类推，$\frac{1}{99×100}$可以写成$\frac{1}{99} - \frac{1}{100}$。由此可得原式$=(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + … + (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{99} - \frac{1}{100})$。去掉括号计算得$1 - \frac{1}{100} = \frac{99}{100}$。
规范解答：
$\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + … + \frac{1}{98×99} + \frac{1}{99×100}$
$=(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + … + (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{99} - \frac{1}{100})$
$=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + … + \frac{1}{98} - \frac{1}{99} + \frac{1}{99} - \frac{1}{100}$
$=1 - \frac{1}{100}$
$=\frac{99}{100}$

$\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + … + \frac{1}{98×99} + \frac{1}{99×100}$
$=(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + … + (\frac{1}{98} - \frac{1}{99}) + (\frac{1}{99} - \frac{1}{100})$
$=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + … + \frac{1}{98} - \frac{1}{99} + \frac{1}{99} - \frac{1}{100}$
$=1 - \frac{1}{100}$
$=\frac{99}{100}$

1.计算：$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots + \frac{1}{380}$。

$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \dots + \frac{1}{380}$
$=1- \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \frac{1}{4} - \frac{1}{5} +\dots+ \frac{1}{19} - \frac{1}{20}$
$=1- \frac{1}{20}$
$= \frac{19}{20}$