例① 计算$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$的和是多少？
思路点拨
观察式子发现后一个分数是前一个分数的一半，可以利用数形结合的思想解答，也可以用凑整法解答。
规范解答
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$=\frac{1}{2}+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{16})+(\frac{1}{16}-\frac{1}{32})+(\frac{1}{32}-\frac{1}{64})$
$=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}$
$=1-\frac{1}{64}$
$=\frac{63}{64}$

$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}$
$=\frac{1}{2}+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{16})+(\frac{1}{16}-\frac{1}{32})+(\frac{1}{32}-\frac{1}{64})$
$=\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\frac{1}{8}-\frac{1}{16}+\frac{1}{16}-\frac{1}{32}+\frac{1}{32}-\frac{1}{64}$
$=1-\frac{1}{64}$
$=\frac{63}{64}$
答：和是$\frac{63}{64}$。

1. 巧算：$1\frac{1}{2}+2\frac{1}{4}+3\frac{1}{8}+4\frac{1}{16}+5\frac{1}{32}+6\frac{1}{64}+7\frac{1}{128}$

$1\frac{1}{2}+2\frac{1}{4}+\dots+7\frac{1}{128}=(1+2+\dots+7)+(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{128})=28+(1-\frac{1}{128})=28\frac{127}{128}$

2. $\frac{1}{2}=1-\frac{1}{2},\frac{1}{6}=\frac{1}{2}-\frac{1}{3},\frac{1}{12}=\frac{1}{3}-\frac{1}{4}······$根据规律计算下面的题目。
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+······+\frac{1}{90}$

$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\dots+\frac{1}{90}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{9}-\frac{1}{10}=\frac{9}{10}$