已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+`````+1/(a+2012)(b+2012)

已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+`````+1/(a+2012)(b+2012)
已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+`````+1/(a+2012)(b+2012)

已知|ab-2|与|b-1|互为相反数,试求代数式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+`````+1/(a+2012)(b+2012)
即|ab-2|+|b-1|=0
ab-2=b-1=0
所以b=1
a=2/b=2
所以原式=1/1*2+1/2*3+……+1/2013*2014
=1-1/2+1/2-1/3+……+1/2013-1/2014
=1-1/2014
=2013/2014

1/1*2+1/2*3+……+1/2013*2014
=1-1/2+1/2-1/3+……+1/2013-1/2014
=1-1/2014
=2013/2014