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母题2 简便计算。
$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72}$
图解思路:观察算式发现:每个分数都可以拆分成两个分数的差:
$\frac{1}{2} \rightarrow \frac{1}{1 × 2} \rightarrow 1 - \frac{1}{2}$, $\frac{1}{6} \rightarrow \frac{1}{2 × 3} \rightarrow \frac{1}{2} - \frac{1}{3}$, $\frac{1}{12} \rightarrow \frac{1}{3 × 4} \rightarrow \frac{1}{3} - \frac{1}{4}$, $\dots$,
$\frac{1}{72} \rightarrow \frac{1}{8 × 9} \rightarrow \frac{1}{8} - \frac{1}{9}$
规范解答: $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72}$
$= \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + \dots + \frac{1}{8 × 9}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{8} - \frac{1}{9})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9}$
$= 1 - \frac{1}{9}$
$= \frac{8}{9}$
$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72}$
图解思路:观察算式发现:每个分数都可以拆分成两个分数的差:
$\frac{1}{2} \rightarrow \frac{1}{1 × 2} \rightarrow 1 - \frac{1}{2}$, $\frac{1}{6} \rightarrow \frac{1}{2 × 3} \rightarrow \frac{1}{2} - \frac{1}{3}$, $\frac{1}{12} \rightarrow \frac{1}{3 × 4} \rightarrow \frac{1}{3} - \frac{1}{4}$, $\dots$,
$\frac{1}{72} \rightarrow \frac{1}{8 × 9} \rightarrow \frac{1}{8} - \frac{1}{9}$
规范解答: $\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72}$
$= \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + \dots + \frac{1}{8 × 9}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{8} - \frac{1}{9})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9}$
$= 1 - \frac{1}{9}$
$= \frac{8}{9}$
答案
$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72}$
$= \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + \dots + \frac{1}{8 × 9}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{8} - \frac{1}{9})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9}$
$= 1 - \frac{1}{9}$
$= \frac{8}{9}$
$= \frac{1}{1 × 2} + \frac{1}{2 × 3} + \frac{1}{3 × 4} + \dots + \frac{1}{8 × 9}$
$= (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{8} - \frac{1}{9})$
$= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9}$
$= 1 - \frac{1}{9}$
$= \frac{8}{9}$
计算下面各题。
1. $\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{48×49}+\frac{1}{49×50}$
2. $\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+…+\frac{71}{72}$
1. $\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{48×49}+\frac{1}{49×50}$
2. $\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+…+\frac{71}{72}$
答案
1. $\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{48×49}+\frac{1}{49×50}$
$=(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{48} - \frac{1}{49}) + (\frac{1}{49} - \frac{1}{50})$
$=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{48} - \frac{1}{49} + \frac{1}{49} - \frac{1}{50}$
$=1 - \frac{1}{50}$
$=\frac{49}{50}$
2. $\frac{1}{2} + \frac{5}{6} + \frac{11}{12} + \dots + \frac{71}{72}$
$=(1 - \frac{1}{2}) + (1 - \frac{1}{6}) + (1 - \frac{1}{12}) + \dots + (1 - \frac{1}{72})$
$=8 - (\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72})$
$=8 - (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9})$
$=8 - (1 - \frac{1}{9})$
$=7 \frac{1}{9}$
$=(1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + \dots + (\frac{1}{48} - \frac{1}{49}) + (\frac{1}{49} - \frac{1}{50})$
$=1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{48} - \frac{1}{49} + \frac{1}{49} - \frac{1}{50}$
$=1 - \frac{1}{50}$
$=\frac{49}{50}$
2. $\frac{1}{2} + \frac{5}{6} + \frac{11}{12} + \dots + \frac{71}{72}$
$=(1 - \frac{1}{2}) + (1 - \frac{1}{6}) + (1 - \frac{1}{12}) + \dots + (1 - \frac{1}{72})$
$=8 - (\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \dots + \frac{1}{72})$
$=8 - (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + \dots + \frac{1}{8} - \frac{1}{9})$
$=8 - (1 - \frac{1}{9})$
$=7 \frac{1}{9}$
