(n^2+n)x^2-(2n+1)x+1=0
由根与系数的关系
x1+x2=(2n+1)/(n^2+n)
x1x2=1/(n^2+n)
|AnBn|=|x1-x2|
=√(x1-x2)^2
=√[(x1+x2)^2-4x1x2]
=√[((2n+1)/(n^2+n))^2-4(n^2+n)]
=√[1/(n^2+n)^2]
=1/(n^2+n)
=1/[n(n+1)]
=1/n-1/(n+1)
因此|A1B1|+...|A2004B2004|
=1/1-1/2+1/2-1/3+...+1/2004-1/2005
=1-1/2005
=2004/2005