已知√a-1+(ab-2)^2=0,求:1/ab +1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2004)(b+2004)

题目

已知√a-1+(ab-2)^2=0,求:1/ab +1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2004)(b+2004)

答案

即a-1=0,ab-2=0
所以a=1
b=2/a=2
所以原式=1/1*2+1/2*3+……+1/2005*2006
=1-1/2+1/2-1/3+……+1/2005-1/2006
=1-1/2006
=2005/2006