《2025年触类旁通奥数点睛六年级数学通用版》

### 金牌例题 1
计算：$\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\cdots+\frac{1}{24\times25}$。

答案： 原式$=1 - \frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\cdots+\frac{1}{24}-\frac{1}{25}=1 - \frac{1}{25}=\frac{24}{25}$

计算。
(1)$\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+\cdots+\frac{1}{59\times60}$
(2)$\frac{1}{2011\times2012}+\frac{1}{2012\times2013}+\frac{1}{2013\times2014}+\frac{1}{2014\times2015}+\cdots+\frac{1}{2021\times2022}$
(3)$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}$
(4)$\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+\cdots+\frac{1}{48\times50}$

答案：
(1)原式$=\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+\cdots+\frac{1}{59}-\frac{1}{60}=\frac{1}{10}-\frac{1}{60}=\frac{1}{12}$
(2)原式$=\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2012}-\frac{1}{2013}+\frac{1}{2013}-\frac{1}{2014}+\frac{1}{2014}-\frac{1}{2015}+\cdots+\frac{1}{2021}-\frac{1}{2022}=\frac{1}{2011}-\frac{1}{2022}=\frac{11}{4066242}$
(3)原式$=\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+\frac{1}{5\times6}+\frac{1}{6\times7}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}=1-\frac{1}{7}=\frac{6}{7}$
(4)原式$=2\times(\frac{1}{2\times4}+\frac{1}{4\times6}+\frac{1}{6\times8}+\cdots+\frac{1}{48\times50})\times\frac{1}{2}=(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+\cdots+\frac{2}{48\times50})\times\frac{1}{2}=(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\cdots+\frac{1}{48}-\frac{1}{50})\times\frac{1}{2}=(\frac{1}{2}-\frac{1}{50})\times\frac{1}{2}=\frac{6}{25}$