1\/1X2+1\/2X3+1\/3X4+...+1\/n(n+1)等于几?
=1-1\/2+1\/2-1\/3+1\/3-1\/4……+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)
一道数学题,请用简便方法计算:1\/1x2+1\/2x3+1\/3x4+...+1\/10x11
10\/11。因为这个式子的每一项都是1\/n(n+1)的形式。且1\/n(n+1)=[(n+1)-n]\/n(n+1)=1\/n-1\/(n+1)所以原式=(1-1\/2)+(1\/2-1\/3)+……+(1\/10-1\/11)=1+(-1\/2+1\/2-1\/3+……+1\/10)-1\/11,括号内都相互抵消 =1-1\/11 =10\/11 乘除法 1、分数乘整数,分母不变...
1\/1x2+1\/2x3+1\/3x4+…+1\/n(n+1)=___,(用含有n的式子表示)
(1) n\/(n+1)(2) n=17 都是裂项求和
1\/1x2+1\/2x3+1\/3x4+...+1\/n(n+1)=?
所以1、1\/1-1\/2+1\/2-1\/3+1\/3-1\/4...1\/n-1\/n+1=1-1\/n+1=n\/n+1 (中间的每2项可以消去)2、2(1\/2-1\/4+1\/4-1\/6...1\/2010-1\/2012)=2*(1\/2-1\/2012)=1-1\/1006=1005\/1006
1\/1x2+1\/2x3+1\/3x4+...+1\/nx(n+1)=__
裂项相消法。1\/(1*2)+1\/(2*3)+1\/(3*4)+...+1\/[n(n+1)]=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+...+[1\/n-1\/(n+1)]=1-1\/(n+1)=n\/(n+1)填:n\/(n+1)
1\/1x2+1\/2x3+1\/3x4+...+1\/nx(n+1)=__
裂项相消法.1\/(1*2)+1\/(2*3)+1\/(3*4)+.+1\/[n(n+1)]=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+.+[1\/n-1\/(n+1)]=1-1\/(n+1)=n\/(n+1)填:n\/(n+1)
求和:1\/1x2+1\/2x3+1\/3x4+···+1\/n(n+1)
1\/1x2+1\/2x3+1\/3x4+···+1\/n(n+1)=1\/2+1\/6+1\/12+...+1\/n(n+1)=1-1\/2+(1\/2-1\/3)+(1\/3-1\/4)+...+(1\/n-1\/n+1)=1-1\/2+1\/2-1\/3+1\/3-1\/4+...+1\/n-1\/n+1 =1-1\/n+1 =n\/n+1
求和1\/(1x2)+1\/(2x3)+1\/(3x4)+……1\/[nx(n+1)]=?
1\/k(k+1)=1\/k-1\/(k+1)所以原式=1\/1-1\/2+1\/2-1\/3...+1\/n-1\/(n+1)=1-1\/(n+1)=n\/(n+1)
1\/1x2十1\/2x3+1\/3x4十…十1\/n(n+1)=?
12
1\/1X2+1\/2x3+1\/3x4+1\/4x5+...+1\/98x99+1\/99x100求结果!!!
1\/1乘2等于1\/1减1\/2 同理2乘3分之一等于2分之一减去3分之一 上式等于1\/1-1\/2+1\/2--1\/3+1\/3...-1\/99+1\/99-1\/100等于99\/100