已知（ab-2）²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)=

已知（ab-2）²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)=
已知（ab-2）²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)=

已知（ab-2）²+|b-1|=0,求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)=
（ab-2）²+|b-1|=0可知,b=1,a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…1/(a+2011)(b+2011)
=1/2+1/(2*3)+……+1/(2013*2012)
=1-1/2+1/2-1/3+1/3-1/4+……+1/1012-1/2013
=2012/1013