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八年级数学填空题一般
题目
类比推理是根据一类事物所具有的某种属性,推测与其类似的事物也具有这种属性的一种推理方法.著名数学家波利亚认为"类比就是一种形似".类比推理思想在初中代数推理学习中也被广泛应用.
【特例感知】
观察下列等式:11×2=1112\frac{1}{1×2}=\frac{1}{1}-\frac{1}{2},12×3=1213\frac{1}{2×3}=\frac{1}{2}-\frac{1}{3}.
(1)(1)根据上述特征,计算:11×2+12×3+13×4+14×5=______.\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}= \_\_\_\_\_\_.
【尝试类比】
(2)(2)已知一次函数y=m+2mx+2m(my=-\frac{m+2}{m}x+\frac{2}{m}(m为正整数)与xx轴、yy轴分别交于AA,BB两点,OO为坐标原点,设RtAOBRt\triangle AOB的面积为SmS_{m}.
S2=S_{2}=______;
②求S2+S4+S6++S2024S_{2}+S_{4}+S_{6}+\cdots +S_{2024}的值.
【类比迁移】
(3)(3)计算:11+2+11+2+3+11+2+3+4++11+2+3++n=______.\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+⋯+\frac{1}{1+2+3+⋯+n}= \_\_\_\_\_\_.
知识点:面积及等积变换章节:数学竞赛 / 几何

答案与解析

答案

(1)11×2+12×3+13×4+14×5=112+1213+1415=45\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5} =\frac{4}{5}
故答案为:45\frac{4}{5}
(2)(2)①对于y=m+2mx+2m(my=-\frac{m+2}{m}x+\frac{2}{m}(m为正整数),令x=0x=0,则y=2my=\frac{2}{m},令y=0y=0,则x=2m+2x=\frac{2}{m+2}
Sm=12×2m×2m+2=2×1m(m+2)=1m1m+2S_{m}=\frac{1}{2}\times \frac{2}{m}\times \frac{2}{m+2}=2\times \frac{1}{m(m+2)}=\frac{1}{m}-\frac{1}{m+2}
m=2m=2时,则S2=1214=14S_{2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}
故答案为:14\frac{1}{4}
②由①知Sm=1m1m+2S_{m}=\frac{1}{m}-\frac{1}{m+2}
S2+S4+S6++S2024=(1214+1416+..S_{2}+S_{4}+S_{6}+\cdots +S_{2024}=(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...1202412026)=1212026=5061013\frac{1}{2024}-\frac{1}{2026})=\frac{1}{2}-\frac{1}{2026}=\frac{506}{1013}
(3)1+2+3+...+n=12n(n+1)(3)\because 1+2+3+...+n=\frac{1}{2}n\left(n+1\right)
11+2+3+...+n=2n(n+1)\frac{1}{1+2+3+...+n}=\frac{2}{n(n+1)}
11+2+11+2+3+11+2+3+4++11+2+3++n=2(1213+1314+14+...+1n1n+1)=2(121n+1)=12n+1\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+⋯+\frac{1}{1+2+3+⋯+n}= 2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1})=2(\frac{1}{2}-\frac{1}{n+1})=1-\frac{2}{n+1}
故答案为:12n+11-\frac{2}{n+1}.

解析

(1)11×2+12×3+13×4+14×5=112+1213+1415=45\frac{1}{1×2}+\frac{1}{2×3}+\frac{1}{3×4}+\frac{1}{4×5}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5} =\frac{4}{5}
故答案为:45\frac{4}{5}
(2)(2)①对于y=m+2mx+2m(my=-\frac{m+2}{m}x+\frac{2}{m}(m为正整数),令x=0x=0,则y=2my=\frac{2}{m},令y=0y=0,则x=2m+2x=\frac{2}{m+2}
Sm=12×2m×2m+2=2×1m(m+2)=1m1m+2S_{m}=\frac{1}{2}\times \frac{2}{m}\times \frac{2}{m+2}=2\times \frac{1}{m(m+2)}=\frac{1}{m}-\frac{1}{m+2}
m=2m=2时,则S2=1214=14S_{2}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}
故答案为:14\frac{1}{4}
②由①知Sm=1m1m+2S_{m}=\frac{1}{m}-\frac{1}{m+2}
S2+S4+S6++S2024=(1214+1416+..S_{2}+S_{4}+S_{6}+\cdots +S_{2024}=(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...1202412026)=1212026=5061013\frac{1}{2024}-\frac{1}{2026})=\frac{1}{2}-\frac{1}{2026}=\frac{506}{1013}
(3)1+2+3+...+n=12n(n+1)(3)\because 1+2+3+...+n=\frac{1}{2}n\left(n+1\right)
11+2+3+...+n=2n(n+1)\frac{1}{1+2+3+...+n}=\frac{2}{n(n+1)}
11+2+11+2+3+11+2+3+4++11+2+3++n=2(1213+1314+14+...+1n1n+1)=2(121n+1)=12n+1\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+⋯+\frac{1}{1+2+3+⋯+n}= 2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1})=2(\frac{1}{2}-\frac{1}{n+1})=1-\frac{2}{n+1}
故答案为:12n+11-\frac{2}{n+1}.

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