Question

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+2009)(b+2009)的值。

Answer 1

|ab-2|与|b-1|互为相反数
∴|ab-2|+|b-1|=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+……+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+……+1/2010-1/2011
=1-1/2011
=2010/2011 猴岛外宣Iim　　 温眸-为你解答

Answer 2

|ab-2|与|b-1|互为相反数
∴|ab-2|+|b-1|=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+……+1/2010×2011
=1-1/2+1/2-1/3+1/3-1/4+……+1/2010-1/2011
=1-1/2011
=2010/2011