求1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.+1/(2005x2006)

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求1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.+1/(2005x2006)

求1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.+1/(2005x2006)
求1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.+1/(2005x2006)

求1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.+1/(2005x2006)
除了最左边和最右边的一项其余通通可以抵消即原式等于1减2006分之1等于2OO6分之2005 我是数学老师不懂再来问我好了.


原式=(2 -1)/(1x2)+(3 -2)/(2x3)+ (4 -3)/(3x4)+ (5 -4)/(4x5)+.....+(2006 -2005)/(2005x2006)
=1-1/2+1/2-1/3+3/1-1/4...+1/2005-1/2006
=1- 1/2006
= 2005/2006

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

1/(n*(n+1)) = 1/n-1/(n+1),这个叫“裂项法”
所以1/(1x2)+1/(2x3)+1/(3x4)+1/(4x5)+.....+1/(2005x2006)=(1-1/2)+(1/2-1/3)+...+(1/2005-1/2006)=2005/2006

原是化简 1-1/2+1/2+1/3+·····+1/2005-1/2006=1-1/2006=2005/2006