|已知|ab-2|与(b-1)2互为相反数,试求式子1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008）(b+2008)的值.

题目

|已知|ab-2|与(b-1)2互为相反数,试求式子1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008）(b+2008)的值.

答案

|ab-2|与(b-1)2互为相反数
ab-2=0
b-1=0
b=1
a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2008）(b+2008)
=-1/a+1/b-1/（a+1）+1/（b+1）-...-1/（a+2008）+1/（b+2008）
=-[1/a+1/a+1+.+1/a+2008]+1/b+1/(b+1)+.+1/(b+2008)
=-1/2-1/3-...-1/2010+1/1+1/2+.+1/2009
=1-1/2010
=2009/2010