Question

一丨。/一,

丨ab-2丨与丨a-1丨互为相反数,求下列式子的值 1/ab+1/(a+1)(b+2)+1/(a+2)(b+2)+..+1/(a+2006)(b+2006)

Answer 1

ab-2=0
a-1=0
∴a=1
b=2
1/ab+1/(a+1)(b+2)+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-……-1/2007+1/2007-1/2008
=1-1/2008
=2007/2008