Question

已知|ab-2|与|b-1|互为相反数，试求下列代数式的值1/ab+1/(a+1)(b+1)+1/

(a+2)(b+2)+......+1/(a+2006)(b+2006)

Answer 1

已知|ab-2|与|b-1|互为相反数，
所以ab-2=0;
b-1=0;
b=1;
a=2;
所以
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+......+1/(a+2006)(b+2006)
=1/1×2+1/2×3+1/3×4+...+1/2007×2008
=1-1/2+1/2-1/3+1/3-1/4+...+1/2007-1/2008
=1-1/2008
=2007/2008;