若|a-1|+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2012)(b+2013)的值.

a-1=0
ab-2=0
∴a=1
b=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.1/(a+2013)(b+2013）
=1/1×2+1/2×3+1/3×4+……+1/2014×2015
=1-1/2+1/2-1/3+1/3-1/4+……+1/2014-1/2015
=1-1/2015
=2014/2015