规律题~~吧

来源：百度知道编辑：UC知道时间：2024/05/24 09:45:06

1/x(x+2)+1/(x+2)(x+4)+1/(x+4)(x+6)+……1/(x+2006)(x+2008)

要过程，谢谢！
提示：1/1*3=1/2(1-1/3)
1/3*5=1/2(1/3-1/5)

=1/2(1/x-1/(x+2)+1/(x+2)-1/(x+4)...+1/(x+2006)-1/(x+2008))
=1/2(1/x-1/(x+2008))
=2007/(2x^2+4016x)

1/x(x+2)=1/2*[1/x - 1/(x+2)]
1/(x+4)(x+2)=1/2*[1/(x+2) - 1/(x+4)]
………………类似的，得：
原式=1/2*[1/x - 1/(x+2008)]
=1004/x(x+2008)

={[(1/x)]-[1/(x+2)]+[1/(x+2)]-[1/(x+4)]+......+[1/(x+2006)]-[1/(x+2008)]}/2
={(1/x)-[1/(x+2008)]}/2
=[2008/x(x-2008)]/2
=1004/x(x-2008)