要仔细观察哦！$\frac{1}{1×3} + \frac{1}{1×5} + \frac{1}{3×5} + \frac{1}{3×7} + \frac{1}{5×7} + \frac{1}{5×9} + \dots + \frac{1}{15×17} + \frac{1}{15×19}$ 像我这样把这些式子分成3组，再观察呢？
这下我会做啦，分组之后就能利用加法交换律和结合律转化为两项裂项来计算了。
我的尝试
$\frac{8}{15} + \frac{12}{105} + \frac{16}{315} + \dots + \frac{32}{3315} + \frac{36}{4845}$
$= \frac{8}{1×3×5} + \frac{12}{3×5×7} + \frac{16}{5×7×9} + \dots + \frac{36}{15×17×19}$
$= \frac{3+5}{1×3×5} + \frac{5+7}{3×5×7} + \frac{7+9}{5×7×9} + \dots + \frac{17+19}{15×17×19}$
$= \frac{1}{1×3} + \frac{1}{1×5} + \frac{1}{3×5} + \frac{1}{3×7} + \frac{1}{5×7} + \frac{1}{5×9} + \dots + \frac{1}{15×17} + \frac{1}{15×19}$
$= ( \frac{1}{1×3} + \dots + ) + ( \frac{1}{1×5} + \dots + ) + ( \frac{1}{3×7} + \dots + )$
$=$
$=$
$=$
$=$

$\frac{8}{15} + \frac{12}{105} + \frac{16}{315} + \dots + \frac{32}{3315} + \frac{36}{4845}$
$= \frac{8}{1×3×5} + \frac{12}{3×5×7} + \frac{16}{5×7×9} + \dots + \frac{36}{15×17×19}$
$= \frac{3+5}{1×3×5} + \frac{5+7}{3×5×7} + \frac{7+9}{5×7×9} + \dots + \frac{17+19}{15×17×19}$
$= \frac{1}{1×3} + \frac{1}{1×5} + \frac{1}{3×5} + \frac{1}{3×7} + \frac{1}{5×7} + \frac{1}{5×9} + \dots + \frac{1}{15×17} + \frac{1}{15×19}$
$= (\frac{1}{1×3}+\frac{1}{3×5}+…+\frac{1}{15×17}) + (\frac{1}{1×5}+\frac{1}{5×9}+…+\frac{1}{13×17}) + (\frac{1}{3×7}+\frac{1}{7×11}+…+\frac{1}{15×19})$
$= \frac{1}{2}×(\frac{2}{1×3}+\frac{2}{3×5}+…+\frac{2}{15×17}) + \frac{1}{4}×(\frac{4}{1×5}+\frac{4}{5×9}+…+\frac{4}{13×17}) + \frac{1}{4}×(\frac{4}{3×7}+\frac{4}{7×11}+…+\frac{4}{15×19})$
$= \frac{1}{2}×(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+…+\frac{1}{15}-\frac{1}{17}) + \frac{1}{4}×(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+…+\frac{1}{13}-\frac{1}{17}) + \frac{1}{4}×(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+…+\frac{1}{15}-\frac{1}{19})$
$= \frac{1}{2}×\frac{16}{17} + \frac{1}{4}×\frac{16}{17} + \frac{1}{4}×\frac{16}{57}$
$= \frac{752}{969}$

计算: $\frac{1}{15}+\frac{1}{105}+\frac{1}{315}+\frac{1}{693}+\frac{1}{1287}+\frac{1}{2145}$。

$=\frac{1}{1×3×5}+\frac{1}{3×5×7}+\frac{1}{5×7×9}+…+\frac{1}{11×13×15}$
$=\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+…+\frac{1}{11×13}-\frac{1}{13×15})$
$=\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{13×15})$
$=\frac{16}{195}$
提示:根据$\frac{1}{15}=\frac{1}{1×3×5}$,$\frac{1}{1×3×5}=\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{3×5})$,将原式变形为$\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+…+\frac{1}{11×13}-\frac{1}{13×15})$,通过加减抵消后还剩下$\frac{1}{4}×(\frac{1}{1×3}-\frac{1}{13×15})$,由此即可算出最终的结果。