1. 计算：$1 - \dfrac{1}{20} - \dfrac{1}{30} - \dfrac{1}{42} - \dfrac{1}{56} - \dfrac{1}{72} - \dfrac{1}{90}$。
答案：1. $ \frac{17}{20} $ 【提示】$ 1 - \frac{1}{20} - \frac{1}{30} - \frac{1}{42} - \frac{1}{56} - \frac{1}{72} - \frac{1}{90} = 1 - (\frac{1}{4} - \frac{1}{5}) - (\frac{1}{5} - \frac{1}{6}) - (\frac{1}{6} - \frac{1}{7}) - (\frac{1}{7} - \frac{1}{8}) - (\frac{1}{8} - \frac{1}{9}) - (\frac{1}{9} - \frac{1}{10}) = 1 - \frac{1}{4} + \frac{1}{5} - \frac{1}{5} + \frac{1}{6} - \frac{1}{6} + \frac{1}{7} - \frac{1}{7} + \frac{1}{8} - \frac{1}{8} + \frac{1}{9} - \frac{1}{9} + \frac{1}{10} = 1 - \frac{1}{4} + \frac{1}{10} = \frac{3}{4} + \frac{1}{10} = \frac{17}{20} $
例 2 计算：$\dfrac{1}{2} + \dfrac{5}{6} + \dfrac{11}{12} + ··· + \dfrac{109}{110}$。
答案：思路分析
观察到每个分数的分子比分母少$1$，分母是两个相邻的自然数的积，可以先把每个分数都看成$1$，然后再减去多算的部分。解答： $\dfrac{1}{2} + \dfrac{5}{6} + \dfrac{11}{12} + ··· + \dfrac{109}{110}$
$= (1 - \dfrac{1}{2}) + (1 - \dfrac{1}{6}) + (1 - \dfrac{1}{12}) + ··· + (1 - \dfrac{1}{110})$
$= 10 - (\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + ··· + \dfrac{1}{110})$
$= 10 - (\dfrac{1}{1 × 2} + \dfrac{1}{2 × 3} + \dfrac{1}{3 × 4} + ··· + \dfrac{1}{10 × 11})$
$= 10 - (1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ··· + \dfrac{1}{10} - \dfrac{1}{11})$
$= 10 - (1 - \dfrac{1}{11})$
$= 10 - 1 + \dfrac{1}{11}$
$= 9\dfrac{1}{11}$
观察到每个分数的分子比分母少$1$，分母是两个相邻的自然数的积，可以先把每个分数都看成$1$，然后再减去多算的部分。解答： $\dfrac{1}{2} + \dfrac{5}{6} + \dfrac{11}{12} + ··· + \dfrac{109}{110}$
$= (1 - \dfrac{1}{2}) + (1 - \dfrac{1}{6}) + (1 - \dfrac{1}{12}) + ··· + (1 - \dfrac{1}{110})$
$= 10 - (\dfrac{1}{2} + \dfrac{1}{6} + \dfrac{1}{12} + ··· + \dfrac{1}{110})$
$= 10 - (\dfrac{1}{1 × 2} + \dfrac{1}{2 × 3} + \dfrac{1}{3 × 4} + ··· + \dfrac{1}{10 × 11})$
$= 10 - (1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + ··· + \dfrac{1}{10} - \dfrac{1}{11})$
$= 10 - (1 - \dfrac{1}{11})$
$= 10 - 1 + \dfrac{1}{11}$
$= 9\dfrac{1}{11}$
2. 计算：$\dfrac{1}{2} + \dfrac{5}{6} + \dfrac{11}{12} + ··· + \dfrac{379}{380}$。
答案：2. $ 18 \frac{1}{20} $ 【提示】$ \frac{1}{2} + \frac{5}{6} + \frac{11}{12} + ··· + \frac{379}{380} = (1 - \frac{1}{2}) + (1 - \frac{1}{6}) + (1 - \frac{1}{12}) + ··· + (1 - \frac{1}{380}) = 19 - (\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + ··· + \frac{1}{380}) = 19 - (\frac{1}{1×2} + \frac{1}{2×3} + \frac{1}{3×4} + ··· + \frac{1}{19×20}) = 19 - (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ··· + \frac{1}{19} - \frac{1}{20}) = 19 - (1 - \frac{1}{20}) = 19 - 1 + \frac{1}{20} = 18 \frac{1}{20} $
例 3 计算：$\dfrac{3}{4} + \dfrac{3}{16} + \dfrac{3}{64} + \dfrac{3}{256} + \dfrac{3}{1024} + \dfrac{3}{4096}$。
答案：思路分析
通过观察，这几个分数后一个数的分母都等于前一个数分母的$4$倍，分子相同，要用“分数的拆分法”来解答。也就是把算式里的各个分数拆分成减法算式，通过加减相互抵消，求得结果。解答： $\dfrac{3}{4} + \dfrac{3}{16} + \dfrac{3}{64} + \dfrac{3}{256} + \dfrac{3}{1024} + \dfrac{3}{4096}$
$= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{64} + \dfrac{1}{128} + \dfrac{1}{256} + \dfrac{1}{512} + \dfrac{1}{1024} + \dfrac{1}{2048} + \dfrac{1}{4096}$
$= 1 - \dfrac{1}{4096}$
$= \dfrac{4095}{4096}$
通过观察，这几个分数后一个数的分母都等于前一个数分母的$4$倍，分子相同，要用“分数的拆分法”来解答。也就是把算式里的各个分数拆分成减法算式，通过加减相互抵消，求得结果。解答： $\dfrac{3}{4} + \dfrac{3}{16} + \dfrac{3}{64} + \dfrac{3}{256} + \dfrac{3}{1024} + \dfrac{3}{4096}$
$= \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{16} + \dfrac{1}{32} + \dfrac{1}{64} + \dfrac{1}{128} + \dfrac{1}{256} + \dfrac{1}{512} + \dfrac{1}{1024} + \dfrac{1}{2048} + \dfrac{1}{4096}$
$= 1 - \dfrac{1}{4096}$
$= \dfrac{4095}{4096}$