=(1/2)*(4-3)/(3*4)+(1/3)*(5-4)/(4*5)+(1/4)*(6-5)/(5*6)+……+(1/98)*(100-99)*(99*100)
=(1/2)*(1/3-1/4)+(1/3)*(1/4-1/5)+(1/4)*(1/5-1/6)+……+(1/98)*(1/99-1/100)
=(1/2)*(1/3)-(1/2)*(1/4)+(1/3)*(1/4)-(1/3)*(1/5)+(1/4)*(1/5)-(1/4)*(1/6)+……+(1/98)*(1/99)-(1/98)*(1/100)
=[(1/2)*(1/3)+(1/3)*(1/4)+(1/4)*(1/5)+......+(1/98)*(1/99)]-[(1/2)(1/4)+(1/3)*(1/5)+(1/4)*(1/6)+......+(1/98)*(1/100)]
=[(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+......+(99-98)/(98*99)]-(1/2)[(4-2)/(2*4)+(5-3)/(3*5)+(6-4)/(4*6)+......+(100-98)/(98*100)]
=[1/2-1/3+1/3-1/4+1/4-1/5+......+1/98-1/99]-(1/2)[1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7.....+1/96-1/98+1/97-1/99+1/98-1/100]
=[1/2-1/99]-(1/2)[1/2-1/99-1/100]
=1/2-1/99-1/4+1/198+1/200
=(1/2-1/99+1/198)+1/200-1/4
=49/99-49/200
=(200*49-99*49)/(99*200
=101*49/(99*200)
=4949/19800
=1/2*{1/[n*(n+1)]-1/[(n+1)(n+2)]}
解：1/(1*2*3)+1/(2*3*4)+1/(3*4*5)+……+1/(98*99*100)
=(1/2)*(4-3)/(3*4)+(1/3)*(5-4)/(4*5)+(1/4)*(6-5)/(5*6)+……+(1/98)*(100-99)*(99*100)
=(1/2)*(1/3-1/4)+(1/3)*(1/4-1/5)+(1/4)*(1/5-1/6)+……+(1/98)*(1/99-1/100)
=(1/2)*(1/3)-(1/2)*(1/4)+(1/3)*(1/4)-(1/3)*(1/5)+(1/4)*(1/5)-(1/4)*(1/6)+……+(1/98)*(1/99)-(1/98)*(1/100)
=[(1/2)*(1/3)+(1/3)*(1/4)+(1/4)*(1/5)+......+(1/98)*(1/99)]-[(1/2)(1/4)+(1/3)*(1/5)+(1/4)*(1/6)+......+(1/98)*(1/100)]
=[(3-2)/(2*3)+(4-3)/(3*4)+(5-4)/(4*5)+......+(99-98)/(98*99)]-(1/2)[(4-2)/(2*4)+(5-3)/(3*5)+(6-4)/(4*6)+......+(100-98)/(98*100)]
=[1/2-1/3+1/3-1/4+1/4-1/5+......+1/98-1/99]-(1/2)[1/2-1/4+1/3-1/5+1/4-1/6+1/5-1/7.....+1/96-1/98+1/97-1/99+1/98-1/100]
=[1/2-1/99]-(1/2)[1/2-1/99-1/100]
=1/2-1/99-1/4+1/198+1/200
=(1/2-1/99+1/198)+1/200-1/4
=49/99-49/200
=(200*49-99*49)/(99*200
=101*49/(99*200)
=4949/19800
=1/2*(1/1*2-1/2*3+1/2*3-1/3*4+……+1/98*99-1/99*100)
=1/2*(1/1*2-1/99*100)
=4949/19800本回答被提问者采纳
1\/1*2*3+1\/2*3*4+1\/3*4*5...1\/99*100*101
1\/（1×2×3）=【1\/（1×2）-1\/（2×3）】×1\/2； 1\/（2×3×4）=【1\/（2×3）-1\/（3×4）】×1\/2,；……所以原式=【（1\/（1×2）-1\/（2×3）+1\/（2×3）-1\/（3×4）+……+1\/（99×100）-1\/（100×101）】×1\/2=（1\/2-1\/10100）×1\/2=5049\/20200。
(1\/1×2×3)+(1\/2×3×4)+(1\/3×4×5)+...(1\/98×99×100)等于多少啊...
原式=1\/2*[2\/（1*2*3）+2\/（2*3*4）+...+2\/(98*99*100)]=1\/2*[(3-1)\/（1*2*3）+(4-2)\/（2*3*4）+...+(100-98)\/(98*99*100)]=1\/2*[3\/（1*2*3）-1\/（1*2*3）+4\/（2*3*4）-2\/（2*3*4）+...+100\/(98*99*100)-98\/(98*99*100)]=1\/2*[...
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\/2-1\/3+1\/3-1\/4+...+1\/99-1\/100 =1-1\/100 =99\/100
1×1\/2+2×1\/3+3×1\/4+5×1\/4+,加99×1\/100等于多少?
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。
如何简便计算:1\/1*2+1\/2*3+1\/3*4+.1\/98*99+1\/99*100
要用软件计算么？另外，可变为：(1\/1+1\/1)+(2\/2+1\/2)+(3\/3+1\/3)+(4\/4+1\/4)+...+(99\/99+1\/99)= 99+(1\/1+1\/2+1\/3+1\/4+...+1\/99)结果是：104.1774
分数巧算:1\/1*2 1\/2*3 +1\/3*4 ……+1\/99*100=( )怎么算?
1+1\/2+1\/3+1\/4+...+1\/100=？告诉你一公式：1\/[n*(n+1)]=1\/n - 1\/(n+1)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乘2分之一加2乘3分之一加3乘4分之一一直加到99乘100分之一等于多少
运用裂项公式 分母是两个连续自然数的乘积的时候，有这样的规律。公式算法如下：1\/1*2+1\/2*3+1\/3*4+...+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\/99×100怎么算
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)把每一项都拆开来，前后抵消，最后只剩下1-1\/100=99\/100
...1加2乘3乘4分之1加3乘4乘5分之一...加98乘99乘100分之1
1\/(98x99x100)=1\/2(1\/98+1\/100)-1\/99 因此可以得到 1\/(1x2x3)+ 1\/(2x3x4) +1\/(3x4x5) +...+ x1\/(98x99x100)=1\/2x(1+1\/2+...+1\/98+1\/3+1\/4+...+1\/100)-(1\/2+1\/3+..+1\/99)=1\/2x(1\/2+1\/2+...+1\/98+1\/99-1\/99+1\/2+1\/3+1\/4+...+1\/99+...
1×1\/2+2×1\/3+3×1\/4+1直加到99×1\/100怎么计算?
=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-...-1\/99+1\/99-1\/100 =1-1\/100 =99\/100 分数计算方法：分数的乘法法则：分数乘分数，用分子相乘做积的分子，分母相乘做积的分母，能约分的先约分。通分的步骤：1、先求出原来几个分数（式）的分母的最简公分母。2、根据分数（式）的基本性质，把...
