若丨ab-2丨+(b-1)的平方=0,求1/ab+1/(a+1) (b+1)+1/(a+2) (b+2)+…+1/(a+2009) (b+2009)的值

问题描述:

若丨ab-2丨+(b-1)的平方=0,求1/ab+1/(a+1) (b+1)+1/(a+2) (b+2)+…+1/(a+2009) (b+2009)的值

|ab-2|+(b-1)²=0
∴ab-2=0
b-1=0
∴a=2
b=1
∴1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+……+1/(a+2009)(b+2009)
=1/1×2+1/2×3+1/3×4+1/4×5+……+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+……+1/2010-1/2011
=1-1/2011
=2010/2011