已知|ab-2|+|a-1|=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011)(b+2011)的值急

Question

已知|ab-2|+|a-1|=0.求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011)(b+2011)的值
急

Answer 1

由|ab-2|+|a-1|=0得：
ab-2≥0
a-1≥0
故只有
ab-2=0
a-1=0
解得：a=1 b=2
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+.+1/(a+2011)(b+2011)
=1/2+1/(2·3)+1/(3·4)+.+1/(2012·2013)
=(2-1)/2+(3-2)/(2·3)+(4-3)/(3·4)+.+(2013-2012)/(2012·2013)
=1-1/2+(1/2-1/3)+(1/3-1/4)+.+(1/2012-1/2013)
=1-1/2013
=2012/2013