《2025年重点中学招生分班五年真题分类卷数学》

20. (2024陕西JDFZ)$\frac {2×2}{1×3}+\frac {4×4}{3×5}+\frac {6×6}{5×7}+\frac {8×8}{7×9}+\frac {10×10}{9×11}$

答案： 解:原式=$\frac{4}{3}+\frac{16}{15}+\frac{36}{35}+\frac{64}{63}+\frac{100}{99}$
=5+$\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+\frac{1}{99}$
=5+$\frac{1}{1×3}+\frac{1}{3×5}+\frac{1}{5×7}+\frac{1}{7×9}+\frac{1}{9×11}$
=5+$\frac{1}{2}$×(1-$\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}$)
=5+$\frac{5}{11}$
=5$\frac{5}{11}$

21. (2024四川BS中学)$\frac {1}{1×2×3}+\frac {1}{2×3×4}+\frac {1}{3×4×5}+... +\frac {1}{98×99×100}$

答案： 解:原式=$\frac{1}{2}×(\frac{1}{1×2}-\frac{1}{2×3})+\frac{1}{2}×(\frac{1}{2×3}-\frac{1}{3×4})+\frac{1}{2}×(\frac{1}{3×4}-\frac{1}{4×5})+…+\frac{1}{2}×(\frac{1}{98×99}-\frac{1}{99×100})$
=$\frac{1}{2}$×($\frac{1}{1×2}-\frac{1}{2×3}+\frac{1}{2×3}-\frac{1}{3×4}+…+\frac{1}{98×99}-\frac{1}{99×100}$)
=$\frac{1}{2}$×($\frac{1}{1×2}-\frac{1}{99×100}$)
=$\frac{4949}{19800}$

22. (2023陕西GDFZ)$\frac {29}{1×3×5}+\frac {27}{3×5×7}+\frac {25}{5×7×9}+... +\frac {1}{29×31×33}$

答案： 解:原式=($\frac{30}{1×3×5}+\frac{30}{3×5×7}+\frac{30}{5×7×9}+…+\frac{30}{29×31×33}$)-($\frac{1}{1×3×5}+\frac{3}{3×5×7}+\frac{5}{5×7×9}+…+\frac{29}{29×31×33}$)
=$\frac{15}{2}$×($\frac{4}{1×3×5}+\frac{4}{3×5×7}+\frac{4}{5×7×9}+…+\frac{4}{29×31×33}$)-($\frac{1}{3×5}+\frac{1}{5×7}+\frac{1}{7×9}+…+\frac{1}{31×33}$)
=$\frac{15}{2}$×($\frac{1}{1×3}-\frac{1}{3×5}+\frac{1}{3×5}-\frac{1}{5×7}+\frac{1}{5×7}-\frac{1}{7×9}+…+\frac{1}{29×31}-\frac{1}{31×33}$)-$\frac{1}{2}$×($\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+…+\frac{1}{31}-\frac{1}{33}$)
=$\frac{15}{2}$×($\frac{1}{3}-\frac{1}{31×33}$)-$\frac{1}{2}$×($\frac{1}{3}-\frac{1}{33}$)
=$\frac{2395}{1023}$

23. (2023陕西TYZ)$\frac {1^{2}+2^{2}}{1×2}+\frac {2^{2}+3^{2}}{2×3}+\frac {3^{2}+4^{2}}{3×4}+... +\frac {9^{2}+10^{2}}{9×10}$

答案： 解:原式=$\frac{2}{1}+\frac{1}{2}+\frac{3}{2}+\frac{2}{3}+\frac{4}{3}+\frac{3}{4}+…+\frac{10}{9}+\frac{9}{10}$
=2+2+2+2+2+2+2+2+2+$\frac{9}{10}$
=18$\frac{9}{10}$

24. (2024陕西TYZ)$\frac {1}{1}+\frac {1}{1+2}+\frac {1}{1+2+3}+... +\frac {1}{1+2+3+... +100}$

答案： 解:原式=1+$\frac{1}{(1+2)×2÷2}+\frac{1}{(1+3)×3÷2}+…+\frac{1}{(1+100)×100÷2}$
=1+$\frac{2}{(1+2)×2}+\frac{2}{(1+3)×3}+…+\frac{2}{(1+100)×100}$
=1+2×($\frac{1}{2×3}+\frac{1}{3×4}+…+\frac{1}{100×101}$)
=1+2×($\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+…+\frac{1}{100}-\frac{1}{101}$)
=1+2×($\frac{1}{2}-\frac{1}{101}$)
=1+1-$\frac{2}{101}$
=1$\frac{99}{101}$

25. (2022陕西GXYZ)$24×(\frac {1}{2×3}+\frac {1}{4×5}+... +\frac {1}{20×21})-(\frac {1}{1^{2}}+\frac {1}{1^{2}+2^{2}}+... +\frac {1}{1^{2}+2^{2}+... +10^{2}})$

答案： 解:原式=24×($\frac{1}{2×3}+\frac{1}{4×5}+…+\frac{1}{20×21}$)-6×($\frac{1}{1×2×3}+\frac{1}{2×3×5}+…+\frac{1}{10×11×21}$)
=24×($\frac{1}{2×3}+\frac{1}{4×5}+…+\frac{1}{20×21}$)-24×($\frac{1}{2×4×3}+\frac{1}{4×6×5}+…+\frac{1}{20×22×21}$)
=24×[($\frac{1}{2×3}-\frac{1}{2×4×3}$)+($\frac{1}{4×5}-\frac{1}{4×6×5}$)+…+($\frac{1}{20×21}-\frac{1}{20×22×21}$)]
=24×($\frac{1}{2×4}+\frac{1}{4×6}+…+\frac{1}{20×22}$)
=24×$\frac{1}{2}$×($\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+…+\frac{1}{20}-\frac{1}{22}$)
=12×($\frac{1}{2}-\frac{1}{22}$)
=$\frac{60}{11}$