4a^2-4ab=2b^2-4b=4=0 ,求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+ ··· +1/(a+2009)(b+2009)

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4a^2-4ab=2b^2-4b=4=0 ,求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+ ··· +1/(a+2009)(b+2009)
4a^2-4ab=2b^2-4b=4=0 ,求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+ ··· +1/(a+2009)(b+2009)

4a^2-4ab=2b^2-4b=4=0 ,求1/ab+1/(a+b)(b+1)+1/(a+2)(b+2)+ ··· +1/(a+2009)(b+2009)
题目有误!应该是:4a-4ab+2b-4b+4=0 (4a-4ab+b)+(b-4b+4)=0 即:(2a-b)+(b-2)=0 解得:a=1;b=2 那么:1/(ab)+1/[(a+1)(b+1)]+……+1/[(a+2009)(b+2009)] =1/2+1/6+……+1/(2010x2011) =(1-1/2)+(1/2-1/3)+……+(1/2010-1/2011) =1-1/2011 =2010/2011