《2025年同步练习册配套检测卷六年级数学上册鲁教版五四制》

24. (14 分)观察下列等式的规律,解答下列问题：
① $\frac{1}{2×4} = \frac{1}{2}×(\frac{1}{2} - \frac{1}{4})$；
② $\frac{1}{4×6} = \frac{1}{2}×(\frac{1}{4} - \frac{1}{6})$；
③ $\frac{1}{6×8} = \frac{1}{2}×(\frac{1}{6} - \frac{1}{8})$
…
(1)按以上规律,第④个等式为 $\frac{1}{8×10} = $ $\frac{1}{2}×(\frac{1}{8}-\frac{1}{10})$；第 $n$ 个等式为 $\frac{1}{2n(2n + 2)} = $ $\frac{1}{2}×(\frac{1}{2n}-\frac{1}{2n + 2})$(用含 $n$ 的代数式表示,$n$ 为正整数)。
(2)按此规律,计算：$\frac{1}{2×4} + \frac{1}{4×6} + \frac{1}{6×8} + \frac{1}{8×10} + \frac{1}{10×12}$。
$\begin{aligned}&\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\frac{1}{8×10}+\frac{1}{10×12}\\=&\frac{1}{2}×(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}×(\frac{1}{4}-\frac{1}{6})+\frac{1}{2}×(\frac{1}{6}-\frac{1}{8})+\frac{1}{2}×(\frac{1}{8}-\frac{1}{10})+\frac{1}{2}×(\frac{1}{10}-\frac{1}{12})\\=&\frac{1}{2}×[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{10})+(\frac{1}{10}-\frac{1}{12})]\\=&\frac{1}{2}×(\frac{1}{2}-\frac{1}{12})\\=&\frac{1}{2}×\frac{5}{12}\\=&\frac{5}{24}\end{aligned}$
(3)探究计算：$\frac{2}{2×5} + \frac{2}{5×8} + \frac{2}{8×11} + … + \frac{2}{2021×2024}$。
$\begin{aligned}&\frac{2}{2×5}+\frac{2}{5×8}+\frac{2}{8×11}+\cdots+\frac{2}{2021×2024}\\=&\frac{2}{3}×(\frac{3}{2×5}+\frac{3}{5×8}+\frac{3}{8×11}+\cdots+\frac{3}{2021×2024})\\=&\frac{2}{3}×[(\frac{1}{2}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11})+\cdots+(\frac{1}{2021}-\frac{1}{2024})]\\=&\frac{2}{3}×(\frac{1}{2}-\frac{1}{2024})\\=&\frac{2}{3}×\frac{1011}{2024}\\=&\frac{337}{1012}\end{aligned}$

答案：
(1)
第④个等式：$\frac{1}{8×10}=\frac{1}{2}×(\frac{1}{8}-\frac{1}{10})$；
第$n$个等式：$\frac{1}{2n(2n + 2)}=\frac{1}{2}×(\frac{1}{2n}-\frac{1}{2n + 2})$。
(2)
$\begin{aligned}&\frac{1}{2×4}+\frac{1}{4×6}+\frac{1}{6×8}+\frac{1}{8×10}+\frac{1}{10×12}\\=&\frac{1}{2}×(\frac{1}{2}-\frac{1}{4})+\frac{1}{2}×(\frac{1}{4}-\frac{1}{6})+\frac{1}{2}×(\frac{1}{6}-\frac{1}{8})+\frac{1}{2}×(\frac{1}{8}-\frac{1}{10})+\frac{1}{2}×(\frac{1}{10}-\frac{1}{12})\\=&\frac{1}{2}×[(\frac{1}{2}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{6})+(\frac{1}{6}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{10})+(\frac{1}{10}-\frac{1}{12})]\\=&\frac{1}{2}×(\frac{1}{2}-\frac{1}{12})\\=&\frac{1}{2}×\frac{5}{12}\\=&\frac{5}{24}\end{aligned}$
(3)
$\begin{aligned}&\frac{2}{2×5}+\frac{2}{5×8}+\frac{2}{8×11}+\cdots+\frac{2}{2021×2024}\\=&\frac{2}{3}×(\frac{3}{2×5}+\frac{3}{5×8}+\frac{3}{8×11}+\cdots+\frac{3}{2021×2024})\\=&\frac{2}{3}×[(\frac{1}{2}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{8})+(\frac{1}{8}-\frac{1}{11})+\cdots+(\frac{1}{2021}-\frac{1}{2024})]\\=&\frac{2}{3}×(\frac{1}{2}-\frac{1}{2024})\\=&\frac{2}{3}×\frac{1011}{2024}\\=&\frac{337}{1012}\end{aligned}$