电子课本网 第75页

第75页

信息发布者:
$=(1-\frac {1}{2})+(1-\frac {1}{6})+(1-\frac {1}{12})+...+(1-\frac {1}{110})$
$=1×10-(\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+...+\frac {1}{110})$
$=10-(1-\frac {1}{2}+\frac {1}{2}-\frac {1}{3}+\frac {1}{3}-\frac {1}{4}+...+\frac {1}{10}-\frac {1}{11})$
$=10-(1-\frac {1}{11})$
$=9\frac {1}{11}$
$=\frac {5}{3}×\frac {3}{3×6}+\frac {5}{3}×\frac {3}{6×9}+\frac {5}{3}×\frac {3}{9×12}+...+\frac {5}{3}×\frac {3}{2997×3000}$
$=\frac {5}{3}×(\frac {1}{3}-\frac {1}{6}+\frac {1}{6}-\frac {1}{9}+\frac {1}{9}-\frac {1}{12}+...+\frac {1}{2997}-\frac {1}{3000})$
$=\frac {5}{3}×(\frac {1}{3}-\frac {1}{3000})$
$=\frac {5}{9}-\frac {1}{1800}$
$=\frac {111}{200}$
$=1-\frac {1}{3}+\frac {1}{3}-\frac {1}{5}+\frac {1}{5}-\frac {1}{7}+\frac {1}{7}-\frac {1}{9}$
$=1-\frac {1}{9}$
$=\frac {8}{9}$