电子课本网 第13页

第13页

信息发布者:
$ =(1-\frac{1}{2})+(1-\frac {1}{6})+(1-\frac {1}{12})+(1-\frac {1}{20})$
$+...+(1-\frac {1}{72})$
$=8-(\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+...+\frac {1}{72})$
$=8-(1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+\frac {1}{4}-\frac {1}{5}$
$+...+\frac {1}{8}-\frac {1}{9})$
$=8-\frac {8}{9}$
$=7 \frac {1}{9}$
$ \frac {2}{63}$
$ \frac {1}{63}$
13
$ \frac {2}{143}$
$ \frac {1}{11}$
$ \frac {1}{13}$
$ \frac {1}{143}$
$=1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+...+\frac {1}{2023}-\frac {1}{2024}$
$=1-\frac {1}{2024}$
$=\frac {2023}{2024}$
$=\frac{1}{2}×(\frac{1}{3}-\frac{1}{5}+\frac {1}{5}-\frac {1}{7}+\frac {1}{7}-\frac {1}{9}+...+\frac {1}{97}-\frac {1}{99})$
$=\frac{1}{2}×\frac {32}{99}$
$=\frac {16}{99}$