已知|ab-2|与|b-1|互为相反数,请求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+1999)(b+1999)的值

已知|ab-2|与|b-1|互为相反数,请求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+1999)(b+1999)的值

互为相反数则|ab-2|+|b-1|=0
所以ab-2=0,b-1=0
b=1,a=2/b=2
所以原式=1/1*2+1/2*3+……+1/2000*2001
=1-1/2+1/2-1/3+……+1/2000-1/2001
=1-1/2001
=2000/2001