一道数学题 1x2\/1+2x3\/1+3x4\/1+4x5\/1+…+39x40\/1
1\/39*40=1\/39-1\/40 ∴1\/1*2+1\/2*3+1\/3*4+1\/4*5+...1\/39*40 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-...1\/39+1\/39-1\/40 =1-1\/40 =39\/40
1x2\/1+2x3\/1+3x4\/1+4x5\/1+…+39x40\/1
1\/1*2=1-1\/21\/2*3=1\/2-1\/31\/3*4=1\/3-1\/4.1\/39*40=1\/39-1\/40∴1\/1*2+1\/2*3+1\/3*4+1\/4*5+.1\/39*40=1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/5-.1\/39+1\/39-1\/40=1-1\/40=39\/40
1x2\/1+2x3\/1+3x4\/1+4x5\/1.19x20\/1
=2x(2-1)+3x(3-1)+4x(4-1)+.+20x(20-1)=2^2+3^2+4^2+.+20^2-(2+3+4+.+20)因为1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)\/6 所以=2869-209 =2660
巧算1x2分之一+2x3分之一+3x4分之一+4x5分之一+...+199x200分之一等于...
1x2分之一+2x3分之一+3x4分之一+4x5分之一+...+199x200分之一 =1-1\/2+1\/2-1\/3+1\/3-...-1\/200 =1-1\/200 =199\/200
...二乘三分一加三乘四分一加四乘五加……加三十九乘四十分之一为什么等...
=41-(1+1\/2+1\/3+...+1\/40)=41-37.69706671716118 = 3.30293328283882 如果题目是 1\/(1*2)+1\/(2*3)+...+1\/(39*40)因为1\/[n*(n+1)]=1\/n-1\/(n+1).故1\/(1*2)+1\/(2*3)+...+1\/(39*40)=1-1\/2+1\/2-1\/3+...+1\/39-1\/40 =1-1\/40 =39\/40 ...
1x2\/1+2x3\/1+3x4\/1+4x5\/1+...23x24\/1 =一乘二分之一...
1x2\/1+2x3\/1+3x4\/1+4x5\/1+...23x24\/1 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+……+1\/23-1\/24 =1-1\/24 =23\/24
1X2分之一+2X3分之1+3X4分之一+4X5分之一+...+99X100分之一用简便运算...
1\/(1*2)+1\/(2*3)+1\/(3*4)+……+1\/(98*99)+1\/(99*100)=(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+……+(1\/98-1\/99)+(1\/99-1\/100)=1-1\/100=99\/100
1×1\/2加2×3分之1+3x4分之一十4x5⃣分之一
1×1\/2加2×3分之1+3x4分之一十4x5⃣分之一 =1-2分之1+2分之1-3分之1+3分之1-4分之1+4分之1-5分之1 =1-5分之1 =5分之4
1x2分之1加2x3分之1加3x4分之1加4x5分之1+…+2013x2014分之1
裂项法:1x2分之1加2x3分之1加3x4分之1加4x5分之1+…+2013x2014分之1 =1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+1\/2013-1\/2014 =1-1\/2014 =2013\/2014
1x2分之1加2x3分之1加3x4分之1加4x5之一加5×6分之一的简便计算
例如1\/(2*3)=1\/2-1\/3 所以原式=1-1\/2+1\/2-1\/3+1\/3+...+1\/5-1\/6=1-1\/6=5\/6