《2025年经纶学典5星学霸七年级数学上册苏科版》

1. 新方法(2024·苏州校级月考)类比推理是一种重要的推理方法,根据两种事物在某些特征上相似,得出它们在其他特征上也可能相似的结论.阅读感知:在异分母分数的加减法中,往往先化作同分母,然后分子相加减,例如:$\frac {1}{2}-\frac {1}{3}= \frac {3}{2×3}-\frac {2}{3×2}= \frac {3-2}{6}= \frac {1}{6}$,我们将上述计算过程倒过来,得到$\frac {1}{6}= \frac {1}{2×3}= \frac {1}{2}-\frac {1}{3}$,这一恒等变形过程在数学中叫作裂项.类似地,对于$\frac {1}{4×6}可以用裂项的方法变形为\frac {1}{4×6}= \frac {1}{2}×(\frac {1}{4}-\frac {1}{6})$.类比上述方法,解决以下问题.
【类比探究】(1)猜想并写出:$\frac {1}{n×(n+1)}=$$\frac{1}{n}-\frac{1}{n+1}$;
【理解运用】(2)类比裂项的方法,计算:$\frac {1}{1×2}+\frac {1}{2×3}+\frac {1}{3×4}+... +\frac {1}{99×100}$;由(1)得,原式$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{99}-\frac{1}{100})=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}$.
【迁移应用】(3)探究并计算:$\frac {1}{-1×3}+\frac {1}{-3×5}+\frac {1}{-5×7}+\frac {1}{-7×9}+... +\frac {1}{-667×669}$.$\frac{1}{-1× 3}+\frac{1}{-3× 5}+\frac{1}{-5× 7}+\frac{1}{-7× 9}+\cdots +\frac{1}{-667× 669}=-\frac{1}{2}× (\frac{2}{1× 3}+\frac{2}{3× 5}+\frac{2}{5× 7}+\frac{2}{7× 9}+\cdots +\frac{2}{667× 669})=-\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}+\cdots +\frac{1}{667}-\frac{1}{669})=-\frac{1}{2}× (1-\frac{1}{669})=-\frac{1}{2}× \frac{668}{669}=-\frac{334}{669}$.

答案： 1.
(1)$\frac{1}{n}-\frac{1}{n+1}$
(2)由
(1)得,原式$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\cdots +(\frac{1}{99}-\frac{1}{100})=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}$.
(3)$\frac{1}{-1× 3}+\frac{1}{-3× 5}+\frac{1}{-5× 7}+\frac{1}{-7× 9}+\cdots +\frac{1}{-667× 669}=-\frac{1}{2}× (\frac{2}{1× 3}+\frac{2}{3× 5}+\frac{2}{5× 7}+\frac{2}{7× 9}+\cdots +\frac{2}{667× 669})=-\frac{1}{2}× (1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\cdots +\frac{1}{667}-\frac{1}{669})=-\frac{1}{2}× (1-\frac{1}{669})=-\frac{1}{2}× \frac{668}{669}=-\frac{334}{669}$.

2. 计算:
(1)$\frac {1}{-2}+\frac {1}{-6}+\frac {1}{-12}+\frac {1}{-20}+\frac {1}{-30}+\frac {1}{-42}$; (2)$19\frac {1}{10}+20\frac {1}{40}+21\frac {1}{88}+22\frac {1}{154}+23\frac {1}{238}$;
(3)$\frac {4}{3}-\frac {7}{12}+\frac {9}{20}-\frac {11}{30}+\frac {13}{42}-\frac {15}{56}+\frac {17}{72}$; (4)$\frac {1}{1×2×3}+\frac {1}{2×3×4}+... +\frac {1}{98×99×100}$.

答案： 2.
(1)原式$=-(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42})=-(\frac{1}{1× 2}+\frac{1}{2× 3}+\frac{1}{3× 4}+\frac{1}{4× 5}+\frac{1}{5× 6}+\frac{1}{6× 7})=- (1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7})=-(1-\frac{1}{7})=-\frac{6}{7}$.
(2)原式$=19+20+21+22+23+\frac{1}{10}+\frac{1}{40}+\frac{1}{88}+\frac{1}{154}+\frac{1}{238}=(21-2)+(21-1)+21+(21+1)+(21+2)+\frac{1}{2× 5}+\frac{1}{5× 8}+\frac{1}{8× 11}+\frac{1}{11× 14}+\frac{1}{14× 17}=5× 21+\frac{1}{3}× (\frac{1}{2}-\frac{1}{5})+\frac{1}{3}× (\frac{1}{5}-\frac{1}{8})+\frac{1}{3}× (\frac{1}{8}-\frac{1}{11})+\frac{1}{3}× (\frac{1}{11}-\frac{1}{14})+\frac{1}{3}× (\frac{1}{14}-\frac{1}{17})=105+\frac{1}{3}× (\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{17})=105+\frac{1}{3}× (\frac{1}{2}-\frac{1}{17})=105\frac{5}{34}$.
(3)原式$=1+\frac{1}{3}-(\frac{1}{3}+\frac{1}{4})+(\frac{1}{4}+\frac{1}{5})-(\frac{1}{5}+\frac{1}{6})+(\frac{1}{6}+\frac{1}{7})-(\frac{1}{7}+\frac{1}{8})+(\frac{1}{8}+\frac{1}{9})=1+\frac{1}{3}-\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+\frac{1}{5}-\frac{1}{5}-\frac{1}{6}+\frac{1}{6}+\frac{1}{7}-\frac{1}{7}-\frac{1}{8}+\frac{1}{8}+\frac{1}{9}=1+\frac{1}{9}=\frac{10}{9}$.
(4)原式$=(\frac{1}{1× 3}-\frac{1}{2× 3})+(\frac{1}{2× 4}-\frac{1}{3× 4})+\cdots +(\frac{1}{98× 100}-\frac{1}{99× 100})=\frac{1}{1× 3}+\frac{1}{2× 4}+\cdots +\frac{1}{98× 100}-\frac{1}{2× 3}-\frac{1}{3× 4}-\cdots -\frac{1}{99× 100}=\frac{1}{2}× (1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\cdots +\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100})-(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{99}-\frac{1}{100})=\frac{1}{2}× (1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100})-(\frac{1}{2}-\frac{1}{100})=\frac{1}{2}× (\frac{1}{2}-\frac{1}{99}+\frac{1}{100})=\frac{4949}{19800}$.