已知（a-1)^2+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2007)(b+2007)^2为2次方

Question

已知（a-1)^2+|ab-2|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2007)(b+2007)
^2为2次方

Answer 1

因为已知又有平方和绝对值,所以a=1 b=2,所以b=a+1.因此1/ab=1/a(a+1)=[(a+1)-a]/a(a+1)=1/a-1/(a+1).由此可以将原式都解开,变成1/1-1/2+(1/2-1/3)+….+(1/2008-1/2009)＝1-1/2009=2008/2009