如果有理数a、b满足|ab-2|加（1-b)的2次方=0,试求：1/ab+1/(a+1)(b+1)+……+1/（a+2003)(b+2003)的值

ab-2=0,1-b=0
a=2,b=1
1/ab+1/(a+1)(b+1)+……+1/（a+2003)(b+2003)
=1/(2*1)+1/(3*2)+……+1/(2005*2004)
=(1-1/2)+(1/2-1/3)+……+(1/2004-1/2005)
=1-1/2005
=2004/2005