第56页 - 亮点给力提优课时作业本七年级数学江苏版

A

$\frac {2024}{2025}$

$解：因为 1+2+3+...+n= \frac{n(n+1)}{2}，$

$所以 \frac{1}{1+2+3+..+n}=\frac{1}{\frac {n(n+1)}{2}}$

$=\frac{2}{n(n+1)}=\frac{2}{n}－\frac{2}{n+1}，$

$所以原式=1+\frac{2}{2}－\frac{2}{3}+\frac{2}{3}－\frac{2}{4}+...+\frac{2}{100}－\frac{2}{101}$

$=2－\frac{2}{101}$

$=\frac{200}{101}$

（更多请点击查看作业精灵详解）

$解：令A=8\frac {17}{27}+13\frac {12}{17}－5\frac {38}{39}$

$因为17\frac {7}{27}+27\frac {7}{27}－11\frac {37}{39}$

$=16\frac {34}{27}+26\frac {24}{17}－10\frac {76}{39}$

$=2A$

$所以原式=2A÷A=2$

$ \begin{aligned}解：原式&=2³×(1³+2³+3³+… +15³) \\ &=8×14400 \\ &=115200 \\ \end{aligned}$

$解：设原式=S，将原式每个括号内的各项都倒序排列，$

$结果不变$

$即S=\frac {1}{2}+(\frac {2}{3}+\frac {1}{3})+(\frac {3}{4}+\frac {2}{4}+\frac {1}{4})+(\frac {4}{5}+\frac {3}{5}+\frac {2}{5}+\frac {1}{5})$

$+…+(\frac {59}{60}+\frac {58}{60}+…+\frac {3}{60}+\frac {2}{60}+\frac {1}{60})$

$将原序和倒序相加$

$则2S=1+2+3+4+… +59$

$=\frac {(1+59)×59}{2}$

$=30×59$

$所以S=15×59=885$

$解：因为S=4+\frac {4}{5}+\frac {4}{5²}+…+\frac {4}{5^{2024}}①$

$所以\frac {1}{5}S=\frac {4}{5}+\frac {4}{5²}+\frac {4}{5³}+…+\frac {4}{5^{2025}}②$

$①－②得：$

$\frac {4}{5}S=4－\frac {4}{5^{2024}}$

$所以S=5－\frac {1}{5^{2024}}$

$解：设 S=\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+...+\frac{1}{6561}$

$=\frac{1}{3}+\frac{1}{3²}+ \frac{1}{3³}+...+\frac{1}{3^{8}}①，$

$则\frac{1}{3}S=\frac{1}{3²}+\frac{1}{3³}+\frac{1}{3^{4}}+...+\frac{1}{3^{9}}②$

$①－②得：$

$\frac{2}{3}S=\frac{1}{3}－\frac{1}{3^{9}}，$

$所以S=\frac{1}{2}－\frac{1}{2×3^{8}}$

$=\frac{3280}{6561}，$

$即原式=\frac{3280}{6561}$