已知|a-1|+(ab+2)^2=0,求ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1+…(a+2009)(b+2009)/1的值
问题描述：
已知|a-1|+(ab+2)^2=0,求ab/1+(a+1)(b+1)/1+(a+2)(b+2)/1+…(a+2009)(b+2009)/1的值
答
|a-1|+(ab+2)^2=0
所以a-1=0,ab+2=0
a=1,b=2a=2
所以原式=1*2/1+2*3/1+……+2010*2011/1
=1-2/1+2/1-3/1+……+2010/1-2011/1
=1-2011/1
=2011/2010