Question

已知a-1的绝对值+ab-2的绝对值=0，求1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2004)(b+2004)的值

Answer 1

由已知a-1的绝对值+ab-2的绝对值=0
得a-1=0且ab-2=0
所以a=1 ab=2 b=2
设所求式1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)的第n项为An，可以看出，An=1/(a+n-1)(b+n-1)
=1/[ab+(n-1)（a+b）+（n-1）^2]
将a=1 ab=2 b=2代入上式
An=1/[2+3(n-1)+(n-1)^2]
=1/(n-1+2)(n-1+1)
=1/n(n+1)
=1/n-1/(n+1)
所以1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2005)(b+2005)
=1-1/2+1/2-1/3+……+1/2005-1/2006+1/2006-1/2007
=1-1/2007
=2006/2007

Answer 2

|a-1|+|ab-2|=0
所以a-1=0,ab-2=0
a=1
b=2/a=2

所以原式=1/1×2+1/2×3+……+1/2005×2006
=1-1/2+1/2-1/3+……+1/2005-1/2006
=1-1/2006
=2005/2006