请教一道小学数学题,1/1x2+1/2x3+1/3x4+1/4x5+.1/2005x2006怎么算?说明思路.谢谢!

1/1*2 + 1/2*3 + …………+1/2005*2006
=(1-1/2)+(1/2-1/3)+…………(1/2005-1/2006)
=1-1/2006=2005/2006

1/1x2+1/2x3+1/3x4+1/4x5+......1/2005x2006
=2/3+3/2+……2006/2005
=1-1/2006=2005/2006

1/1x2=1-1/2
1/2x3=1/2-1/3
1/3x4=1/3-1/4
依此类推
1/2005x2006=1/2005-1/2006
化简后 中间项全部抵消
原式＝1-1/2006＝2005/2006

1/1*2=1-1/2
1/2*3=1/2-1/3
...
1/2005*2006=1/2005-1/2006
所以
原式=1-1/2-(1/2-1/3)-(1/3-1/4)-...-(1/2005-1/2006)
=1-1/2006
=2005/2006

1/1x2=1/1-1/2 1/2x3=1/2-1/3 1/3x4=1 原式=1/1-1/2+1/2-1/3+1/3-1/4+1/4*** ***-1/2005+1/2005-1/2006=1-1/2006=2005/2006

由1/1=1-1/2
1/2*3=1/2-1/3
......
得1/1x2+1/2x3+1/3x4+1/4x5+......1/2005x2006
=1-1/2+1/2-1/3.......+1/2005-1/2006
=1-1/2006
=2005/2006

我认为你的题目应该为1/(1x2)+1/(2x3)+......1/(2005x2006)
观察每一项,找通项公式,然后分解通项公式,加起来,消去
中间项,只剩下1和-1/2006,步骤如下:
每一项等于1/n*(n+1)=1/n-1/(n+1)
原式=1-1/2+1/2-1/3+1/3-1/4+......+1/2005-1/2006
=1-1/2006=2005/2006

根据两个连续自然数A,B.1/A-1/B=1/AB得原式=1-1/2+(1/2-1/3)+(1/4-1/3)+(1/5-1/4)+...+(1/2006-1/2005)=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5+...+1/2005-1/2006)=1-1/2006=2005/2006.

1-1/2+1/2-1/3.......+1/2005-1/2006
=1-1/2006
=2005/2006

1-1/2+1/2-1/3.......+1/2005-1/2006
=1-1/2006
=2005/2006