1/(1+2+……+n)=2/n(n+1)=2*[1/n-1/(n+1)
所以1+1/1+2+1/1+2+3+……+1/1+2+3+……+100
=2*[(1/1-1/2)+(1/2-1/3)+……+(1/100-1/101)]
=2*(1/1-1/101)
=200/101
1/(1+2+3+……+n)=2/[n*(n+1)]
1+1/(1+2)+1/(1+2+3)+……+1/(1+2+3+……+100)
=1+2/(2*3)+2/(3*4)+……+2/(100*101)
=2*(1-1/2+1/2-1/3+1/3-1/4+……+1/100-1/101)
=2*(1-1/101)
=200/101
1+1\/1+2+1\/1+2+3+……+1\/1+2+3+……+100
所以1+1\/1+2+1\/1+2+3+……+1\/1+2+3+……+100 =2*[(1\/1-1\/2)+(1\/2-1\/3)+……+(1\/100-1\/101)]=2*(1\/1-1\/101)=200\/101
1+(1\/1+2)+(1\/1+2+3)+…+(1\/1+2+3+…+100)的和
即有:1+（1\/1+2）+（1\/1+2+3）+…+（1\/1+2+3+…+100）=1+2(1\/2-1\/3)+2(1\/3-1\/4)+2(1\/4-1\/5)+...+2(1\/100-1\/101)=1+1-2\/3+2\/3-2\/4+2\/4-2\/5+...+2\/100-2\/101 =2-2\/101 =200\/101
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…100)=
1+2=2*3\/2 1+2+3=3*4\/2 1+2+3+4=4*5\/2 1+2+3+……+100=100*101\/2 所以，1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+2006)=1+2\/（2*3）+2\/（3*4）+2\/（4*5）+……+2\/（100*101）=2[（1\/2+1\/（2*3）+1\/（3*4）+1\/（4*5...
奥数题1+1\/1+2+1\/1+2+3+1\/1+2+3+4+……+1\/1+2+3+4+5+……100
推导过程：1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……+1\/(1+2+3……+n)= 1+1\/[(1+2)×2÷2]+1\/[(1+3)×3÷2]+……+1\/[(1+n)×n÷2]——① = 2\/2+2\/(1+2)×2+2\/(1+3)×3+……+2\/(1+n)×n——② = 2×[1\/2+1\/2-1\/3+1\/3-1\/4+……+1...
1+1\/1+2+1\/1+2+3+1\/1+2+3+4+1\/1+2+3+4+5+...+1\/!+2+3+...+100=?_百 ...
1\/1+2+3+...+n=2\/n(n+1)=(2\/n)-(2\/n+1)所以1+ 1\/1 +2+ 1\/1 +2+3+ 1\/1 +2+3+4+ 1\/1 +2+3+4+5+...+ 1\/1 +2+3+...+100=1+(2\/2-2\/3)+(2\/3-2\/4)+(2\/4-2\/5)+...+(2\/100-2\/101)=1+1-2\/101=200\/101 ...
1+1\/1+2+1\/1+2+3+1\/1+2+3+4+.………+1\/1+2+3+.……+10的简便运算
简便运算：1+1\/1+2+1\/1+2+3+……1\/1+2+3+4+5+6+7+8+9+10 （请写出解答过程，OK？） 原式=1+1\/3+1\/6+1\/10+。。。+1\/45+1\/55 =2(1\/2+1\/6+1\/12+1\/20+。。。+1\/90+1\/110) =2(1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+。。。+1\/9-1\/10+1\/10-1...
1+1\/1+2+1\/1+2+3+...+1\/1+2+3+4+...+99+100
1+1\/1+2+1\/1+2+3+1\/1+2+3+4+...1\/1+2+3+...+100 =1+1\/[(1+2)×2÷2]+1\/[(1+3)×3÷2]+1\/[(1+4)×4÷2]+...1\/[(1+100)×100÷2]=1+2\/(1+2)×2+2\/(1+3)×3+2\/(1+4)×4+...2\/(1+100)×100 =1+2×(1\/2-1\/3+1\/3-1\/4+1\/4-1\/5...
1+1\/1+2+1\/1+2+3+1\/1+2+3+4+...+1\/1+2+3+...+100这道数学题怎么做,尽量...
化减：2\/1*2+2\/2*3+2\/3*4+…然后列项：(2\/1-2\/2)+(2\/2-2\/3)…然后化简：2\/1-2\/101=200\/101 如果取极限得2
1+1+2分之1+1+2+3分之1+……+1+2+3+……+100分之1
1+1\/(1+2)+1\/(1+2+3)+……+1\/(1+2+3+……+100)= 2\/(1×2)+2\/(2×3)+2\/(3×4)+……+2\/(100×101)= 2×[1\/(1×2)+1\/(2×3)+1\/(3×4)+……+1\/(100×101)]= 2×[(1-1\/2)+(1\/2-1\/3)+(1\/3-1\/4)+……+(1\/100-1\/101)]= 2×(1-1\/101)= ...
1+ 1\/1+2 + 1\/1+2+3 + 1\/1+2+3+4 +… 1\/1+2+3…+100=
1+1\/3+1\/6+1\/10+1\/15+……+1\/5050 =2(1\/1-1\/2)+2(1\/2-1\/3)+2(1\/3-1\/4)+2(1\/5-1\/6)+...+2(1\/100-1\/101)=2(1\/1-1\/2+1\/2-1\/3+1\/3-1\/4+1\/5-1\/6+...+1\/100-1\/101)=2(1\/1-1\/101)=2*(100\/101)=200\/101 ...
