1 + 1/1+2 + 1/1+2+3 + 1/1+2+3+4 + ...... + 1/1+......+100
谁能教一下这道题目啊，最好讲的详细一点，我能听懂就行，别说太复杂了。
虽然暂时没有悬赏分，但是只要回答的好，就会补充悬赏分15-30
=1+2/2X3+2/3X4+2/4X5+…2/100X101
=1+2(1/2X3+1/3X4+1/4X5+…1/100X101)
=1+2(1/2-1/3+1/3-1/4+1/4-1/5+.....+1/100-1/101)
=1+2(1/2-1/101)
=1+2(99/202)
=200/101
1 + 1\/1+2 + 1\/1+2+3 + 1\/1+2+3+4 + ... + 1\/1+...+100
原式可化为1+1\/3+1\/6+1\/10+1\/15+1\/21+...+1\/5050 =1+2\/2X3+2\/3X4+2\/4X5+…2\/100X101 =1+2(1\/2X3+1\/3X4+1\/4X5+…1\/100X101)=1+2(1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+...+1\/100-1\/101)=1+2(1\/2-1\/101)=1+2(99\/202)=200\/101 ...
数列1,1\/1+2,1\/1+2+3,1\/1+2+3+4 ... 1\/1+2+3+4+...+n,...的前n项的...
所以1\/(1+2+……+n)=2\/n(n+1)=2*[1\/n-1\/(n+1)]所以原式=2*(1\/1-1\/2)+2*(1\/2-1\/3)+2*(1\/3-1\/4)+……+2*[1\/n-1\/(n+1)]中间正负抵消 =2*[1-1\/(n+1)]=2n\/(n+1)
1+1\/1+2+1\/1+2+3+1\/1+2+3+4+……1\/1+2+3+...+100等于多少
答：题目应该缺少了大量的括号吧？1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)第n项的分母是自然数之和(n+1)*n\/2 所以：1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+...+1\/(1+2+3+...+100)=1+2\/(2*3)+2\/(3*4)+2\/(4*5)+...+2\/(...
怎样算出:1+1\/1+2,+1\/1+2+3,+1\/1+2+3+4...1+1\/1+2+3+4+5...+100_百 ...
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+...+100)=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）+……+1\/（100*101）〕因为：...
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 ...
数学题:1\/1+2+1\/1+2+3+1\/1+2+3+4...+1\/1+2+3+...+2001
所以原式有：1\/（1+2）+1\/（1+2+3）+1\/（1+2+3+4）...+1\/（1+2+3+...+2001）=1\/[（1+2）*2\/2]+1\/[（1+3）*3\/2]+1\/[(1+4)*4\/2]+……+1\/[（1+2001）*2001\/2]=2\/(2*3)+2\/(3*4)+2\/4*5+……+2\/（2002*2001）=[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+...+99+100
1+2+3)+1\/(1+2+3+4)+……+1\/(1+2+3+……+100)=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\/5-1\/6)+……+(1\/99-1\/100)]=1+2*(1\/2-1\/100)=1+49\/50 =99\/50 ...
1+1\/(1+2)+1\/(1+2+3)+…+1\/(1+2+3+…100)=
第二种：因为：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*...
计算巧算1+1\/(1+2)+1\/(1+2+3)+1\/(1+2+3+4)+……1\/(1+2+3+……+100...
100*101）〕因为：1\/（2*3）=1\/2-1\/3；1\/（3*4）=1\/3-1\/4；1\/（4*5）=1\/4-1\/5；……1\/（100*101）=1\/2006-1\/101 所以，原式=2（1\/2+1\/2-1\/3+1\/3-1\/4+1\/4-1\/5+……+1\/100-1\/101）=2（1-1\/101）=2*100\/101 =200\/101 检查下吧,可能有打错的......
