《2025年通成学典课时作业本七年级数学上册苏科版宿迁专版》

4. 计算：
(1)$-1-2-3-4-… -199-200$；
(2)$1+5+5^{2}+5^{3}+… +5^{217}+5^{218}$.

答案： （1）设$S=-1-2-3-4-\cdots-199-200$①,则$S=$-200-199-198-197-…-2-1②.由①+②,得$2S=-201×200$,即$2S=-40200$,所以$S=-20100$,即-1-2-3-4-…-199-200=-20100 （2）令$S=1+5+5^{2}+5^{3}+\cdots+5^{217}+5^{218}$①,则$5S=5+5^{2}+5^{3}+5^{4}+\cdots+5^{218}+5^{219}$②.由②-①,得$5S-S=5^{219}-1$,所以$S=\frac{5^{219}-1}{4}$,即$1+5+5^{2}+5^{3}+\cdots+5^{217}+5^{218}=\frac{5^{219}-1}{4}$

5. 计算:$1+\frac {1}{2}+\frac {1}{2^{2}}+\frac {1}{2^{3}}+… +\frac {1}{2^{2026}}.$

答案： 设$S=1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2026}}$①,则$\frac{1}{2}S=\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\frac{1}{2^{4}}+\cdots+\frac{1}{2^{2027}}$②.由①-②,得$\frac{1}{2}S=1-\frac{1}{2^{2027}}$,所以$S=2-\frac{1}{2^{2026}}$,即$1+\frac{1}{2}+\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+\frac{1}{2^{2026}}=2-\frac{1}{2^{2026}}$

6. 阅读材料：
在计算$\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+… +\frac {1}{420}$时,直接计算很繁琐,我们可以采用“裂项——消项”法简化运算.
$\frac {1}{2}+\frac {1}{6}+\frac {1}{12}+\frac {1}{20}+… +\frac {1}{420}= \frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+\frac {1}{4×5}+… +\frac {1}{20×21}= (1-\frac {1}{2})+(\frac {1}{2}-\frac {1}{3})+(\frac {1}{3}-\frac {1}{4})+(\frac {1}{4}-\frac {1}{5})+… +(\frac {1}{20}-\frac {1}{21})= 1-\frac {1}{21}= \frac {20}{21}.$
方法应用：
试用“裂项——消项”法解下面各题：
(1)$\frac {1}{3×7}+\frac {1}{7×11}+\frac {1}{11×15}+… +\frac {1}{55×59}$；
(2)$-\frac {1}{3}-\frac {1}{15}-\frac {1}{35}-\frac {1}{63}-\frac {1}{99}-\frac {1}{143}.$

答案： （1）原式$=\frac{1}{4}×(\frac{1}{3}-\frac{1}{7})+\frac{1}{4}×(\frac{1}{7}-\frac{1}{11})+\frac{1}{4}×(\frac{1}{11}-\frac{1}{15})+\cdots+\frac{1}{4}×(\frac{1}{55}-\frac{1}{59})=\frac{1}{4}×(\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\cdots+\frac{1}{55}-\frac{1}{59})=\frac{1}{4}×(\frac{1}{3}-\frac{1}{59})=\frac{14}{177}$ （2）原式$=-\frac{1}{1×3}-\frac{1}{3×5}-\frac{1}{5×7}-\frac{1}{7×9}-\frac{1}{9×11}-\frac{1}{11×13}=-\frac{1}{2}×(1-\frac{1}{3})-\frac{1}{2}×(\frac{1}{3}-\frac{1}{5})-\frac{1}{2}×(\frac{1}{5}-\frac{1}{7})-\frac{1}{2}×(\frac{1}{7}-\frac{1}{9})-\frac{1}{2}×(\frac{1}{9}-\frac{1}{11})-\frac{1}{2}×(\frac{1}{11}-\frac{1}{13})=-\frac{1}{2}×(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13})=-\frac{1}{2}×(1-\frac{1}{13})=-\frac{6}{13}$