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)

Answer 1

|ab-2|与|b-1|互为相反数.
所以ab-2=0,b-1=0
a=2,b=1

原式=1/(1x2)+1/(2x3)+1/(3x4)+...+1/(2010x2011)
=1-1/2+1/2-1/3+1/3-1/4+...+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 2

|ab-2|与|b-1|互为相反数

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)
=1/2x1+1/3x2+...+1/2011x2010
=1-1/2+1/2-1/3+...+1/2010-1/2011
=1-1/2011
=2010/2011

Answer 3

|ab-2|与|b-1|互为相反数说明|ab-2|=0 |b-1|=0
ab=2 b=1 a=2
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+...+1/(a+2009)(b+2009)
=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