第102页

信息发布者:
​$\frac {1}{⑧}-\frac {1}{⑨}$​
​$=\frac {1}{7×8×9}- \frac {1}{8×9×10}$​
​$=\frac {1}{7}×\frac {1}{8}×\frac {1}{9}-\frac {1}{8}×\frac {1}{9}×\frac {1}{10}$​
​$=\frac {1}{9}×(\frac {1}{7}×\frac {1}{8}-\frac {1}{8}×\frac {1}{10})$​
​$=\frac {1}{9}×[\frac {1}{8}×(\frac {1}{7}-\frac {1}{10}))]$​
​$A=\frac {1}{8}×(\frac {1}{7}-\frac {1}{10})$​
​$=\frac {1}{8}×\frac {3}{70}$​
​$=\frac {3}{560}$​
1
$\frac{1}{64}$
$\frac{63}{64}$

$\begin{aligned}&\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}\\=&\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\frac{1}{128}+\frac{1}{128}-\frac{1}{128}\\=&\frac{1}{4}-\frac{1}{128}\\=&\frac{31}{128}\end{aligned}$

$\begin{aligned}&\frac{7}{5}+\frac{7}{10}+\frac{7}{20}+\frac{7}{40}\\=&\frac{7}{5}+\frac{7}{10}+\frac{7}{20}+\frac{7}{40}+\frac{7}{40}-\frac{7}{40}\\=&\frac{7}{5}\times2-\frac{7}{40}\\=&\frac{21}{8}\end{aligned}$
​$\frac {1}{1×2×3}+\frac {1}{2×3×4}+·s+\frac {1}{98×99×100}=$​
​$(\frac {1}{1×2}-\frac {1}{2×3})×\frac {1}{2}+(\frac {1}{2×3}-\frac {1}{3×4})×\frac {1}{2}+·s+$​
​$(\frac {1}{98×99}-\frac {1}{99×100})×\frac {1}{2}=(\frac {1}{1×2}-\frac {1}{2×3}+$​
​$\frac {1}{2×3}-\frac {1}{3×4}+·s+\frac {1}{98×99}-\frac {1}{99×100})×\frac {1}{2}=$​
​$(\frac {1}{2}-\frac {1}{99×100})×\frac {1}{2}=\frac {4949}{19800}$​