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)
å¡«ï¼n/(n+1)
1\/1x2+1\/2x3+1\/3x4+...+1\/nx(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)
1\/1x2+1\/2x3+1\/3x4+...+1\/n(n+1)=?
由公式 1\/nx（n+m）=m（1\/n-1\/n+m）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\/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)怎么算
1\/1x2+1\/2x3+1\/3x4+...+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)
一道数学题,请用简便方法计算:1\/1x2+1\/2x3+1\/3x4+...+1\/10x11_百度知 ...
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，分子乘整数，最后能约分的要约分。2、分数乘分数，用分子乘分子，用分母乘分母，最后能约分的要约分。3、分数除以...
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
1\/1x2十1\/2x3+1\/3x4十…十1\/n(n+1)=?
12
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)
已知一个数列1\/1X2,1\/2X3,1\/3X4,…1\/n(n+1)…,则前n项的和Sn=?_百度...
此类题最常见的解法是拆项法来解因为1\/n(n+1)=1\/n-1\/(n+1)所以Sn=1\/1X2+1\/2X3+1\/3X4+…+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)求1\/1X3+1\/3X5+1\/5X7,…1\/（2n-1)(2n+1)...
