1. 🌼|新方法（2025·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：$\frac{1}{n×(n+1)}=$；
【理解运用】(2)类比裂项的方法，计算：$\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{99×100}$；
【迁移应用】(3)探究并计算：$\frac{1}{-1×3}+\frac{1}{-3×5}+\frac{1}{-5×7}+\frac{1}{-7×9}+…+\frac{1}{-667×669}$.

1.(1) 因为$\frac{1}{2}=\frac{1}{1×2}=1-\frac{1}{2},\frac{1}{6}=\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3},\frac{1}{12}=\frac{1}{3×4}=\frac{1}{3}-\frac{1}{4}$,类比可得$\frac{1}{n×(n+1)}=\frac{1}{n}-\frac{1}{n+1}$.
(2) 由(1)得，原式$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+\dots+(\frac{1}{99}-\frac{1}{100})=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}$.
(3) 原式$=\frac{1}{-1×3}+\frac{1}{-3×5}+\frac{1}{-5×7}+\frac{1}{-7×9}+\dots+\frac{1}{-667×669}=-\frac{1}{2}×(\frac{2}{1×3}+\frac{2}{3×5}+\frac{2}{5×7}+\frac{2}{7×9}+\dots+\frac{2}{667×669})=-\frac{1}{2}×(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\dots+\frac{1}{667}-\frac{1}{669})=-\frac{1}{2}×(1-\frac{1}{669})=-\frac{1}{2}×\frac{668}{669}=-\frac{334}{669}$.

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)$\frac{1}{1×2×3}+\frac{1}{2×3×4}+\dots+\frac{1}{98×99×100}$.
$\gg$ 进一步挑战进阶专题:P24 专题 10,P25 专题 11

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$=(\frac{1}{1×3}-\frac{1}{2×3})+(\frac{1}{2×4}-\frac{1}{3×4})+\dots+(\frac{1}{98×100}-\frac{1}{99×100})=\frac{1}{1×3}+\frac{1}{2×4}+\dots+\frac{1}{98×100}-\frac{1}{2×3}-\frac{1}{3×4}-\dots-\frac{1}{99×100}=\frac{1}{2}×(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\dots+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100})-(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{99}-\frac{1}{100})=\frac{1}{2}×(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100})-(\frac{1}{2}-\frac{1}{100})=\frac{1}{2}×(\frac{1}{2}-\frac{1}{99}+\frac{1}{100})=\frac{4949}{19800}$.