已知|a-b-1|+|b-1|=0,求1⼀ab+1⼀(a+1)(b+1)+1⼀(a+2)(b+2)+...+1⼀(a+2012)(b+2012)的值

2024-12-19 13:52:09
推荐回答(1个)
回答1:

|a-b-1|+|b-1|=0
∴a-b-1=0
b-1=0
∴a=2
b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2012)(b+2012)
=1/1×2+1/2×3+1/3×4+……+1/2013×2014
=1-1/2+1/2-1/3+1/3-1/4+……+1/2013-1/2014
=1-1/2014
=2013/2014