小学奥数题1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/(99*100)
1\/(1*2)+1\/(2*3)+1\/(3*4)+1\/(4*5)+1\/(5*6)+……+1\/(98*99)+1\/(99*100)=1-1\/2+1\/2-1\/3+...+1\/99-1\/100 =1-1\/100 =99\/100
1\/(1×2)+1\/(2×3)+1\/(3×4)……+1\/(98×99)
1\/(2*3)=(1\/2)-(1\/3)1\/(3*4)=(1\/3)-(1\/4)以此类推 最后答案为1-1\/99=98\/99
1\/(1×2×3)+1\/(2×3×4)+...+1\/(98×99×100)的答案
(1\/98-1\/99)*1\/100 = 1\/98*1\/100-1\/99*1\/100 则原式可变为:1*1\/3+1\/2*1\/4+...1\/98*1\/100 -(1\/2*1\/3+1\/3*1\/4...+1\/99*1\/100)利用裂项求和,分别求两部分的和值:S1=1*1\/3+1\/2*1\/4+...1\/98*1\/100=1\/2*(1-1\/3+1\/2-1\/4+...1\/98-1\/100)=...
1×1\/2加2×1\/3加3×1\/4直到99x1\/100+100乘101分之一
题目应该是1\/(1×2)+1\/(2×3)+1\/(3×4)+……+1\/(99×100)+1\/(100×101),具体如下:1\/(1×2)+1\/(2×3)+1\/(3×4)+……+1\/(99×100)+1\/(100×101)=(1-2分之1)+(2分之1-3分之1)+(3分之1-4分之1)+……+(99分之1-100分之1)+(100分...
1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/99*100怎么用简便方法计算
1\/1*2+1\/2*3+1\/3*4+1\/4*5...+1\/99*100 =1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100
简算。 1.1\/1×2+1\/2×3+1\/3×4+1\/4×5+……+1\/199×200
1\/【1×2】+1\/【2×3】+1\/【3×4】+1\/【4×5】+……+1\/【199×200】=1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/199-1\/200 =1\/1-1\/200 =199\/200
1×1\/2+2×1\/3+3×1\/4+1直加到99×1\/100怎么计算
解答过程如下:1+1\/2+1\/6+...+1\/(99×100)=1+(1-1\/2)+(1\/2-1\/3)+...+(1\/99-1\/100)=2-1\/100 =199\/100
1×1\/2加2×1\/3加3×1\/4加省略号99×1\/100等于多少
99\/100 解题过程如下:1\/1*2+1\/2*3+1\/3*4+...+1\/99*100 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)=1-1\/100 =99\/100
1÷(1×2)+1÷(2×3)+1÷(3×4)……+1÷(99×100)
..+1÷99-1÷100 =1-1÷100 =99÷100 也可以换成分数 1\/1*2+1\/2*3+1\/3*4+...+1\/99*100 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)=11-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100 ...
1×1\/2加,2×1\/3加,3×1\/4,加4×1\/5加到,99×1\/100
1\/1*2+1\/2*3+1\/3*4+...+1\/99*100 =(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/99-1\/100)=1-1\/100 =99\/100 简便计算是一种特殊的计算,它运用了运算定律与数字的基本性质,从而使计算简便,使一个很复杂的式子变得很容易计算出得数。