(1/1×2×3)+(1/2×3×4)+(1/3×4×5)+.......(1/98×99×100)
等于多少啊???
非常着急啊,急啊,那路高手帮个忙啊
(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*3)+1\/(2*3*4)+1\/(3*4*5)+……+1\/(99*100*101)等于多少???
得出的结果是 -1\/2+1\/101 下面算相加的一组,相加的一组要先乘以2,最后再除以2,我们算乘2后,可以变为1-1\/3+1\/2-1\/4+1\/3-1\/5+1\/4-1\/6+……+1\/97-1\/99+1\/98-1\/100+1\/99-1\/101 这样最后还剩余的项有1+1\/2-1\/100-1\/101这四项,因为我们乘过2,所以现在再除以2,得...
1×2×3分之1+2×3×4分之1+3×4×5分之1…… 98×99×100分之1
裂项法:1\/1×2×3+1\/2×3×4+1\/3×4×5+……+1\/98×99×100 =1\/2×(1\/1×2-1\/2×3+1\/2×3-1\/3×4+1\/3×4-1\/4×5+……+1\/98×99-1\/99×100)=1\/2×(1\/1×2-1\/99×100)=1\/2×(1\/2-1\/9900)=1\/2×4949\/9900 =4949\/19800 ...
...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*3)+1\/(2*3*4)+1\/(3*4*5)+...+1\/(9*10*11)简便运算
=1\/2[1\/(1×2)-1\/(2×3)+1\/(2×3)-1\/(3×4)+...+1\/(9×10)-1\/(10×11)]=1\/2×[1\/(1×2)-1\/(10×11)]=1\/2×(1\/2-1\/110)=1\/2×54\/110 =27\/110 公式:1\/n(n+1)(n+2)=1\/2[1\/n(n+1)-1\/(n+1)(n+2)]明教为您解答,如若满意,请点击[满意答案];...
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\/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
1×2+2×3+3×4+...+98×99+99×100=( ? )
+ 3×4×5 - 2×3×4 + …+ 99×100×101-98×99×100)\/3 (可以看出式子中正负相抵消)=99×100×101\/3 =333300 适用于分式形式的通项公式,把一项拆成两个或多个的差的形式,即an=f(n+1)-f(n),然后累加时抵消中间的许多项。常用公式:1、2、3、4、 (当a≠b时)5、...
1\/1×2×3+1\/2×3×4+1\/3×4×5+1\/4×5×6+...+1\/48×49×50的计算过 ...
1\/1×2×3+1\/2×3×4+1\/3×4×5+1\/4×5×6+...+1\/48×49×50 =(1\/2)*(1\/1×2-1\/2×3)+(1\/2)*(1\/2×3-1\/3×4)+...(1\/2)*(1\/48×49-1\/49×50)=(1\/2)*(1\/1×2-1\/2×3+1\/2×3-1\/3×4+1\/3×4...+1\/48×49-1\/49×50)=(1\/2)*(1\/1×2...
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