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+2012)(b+2012)

Answer 1

因为|ab-2|与|b-1|互为相反数
所以ab-2=0 b-1=0
即a=2 b=1
所以
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

Answer 2

|ab-2|与|b-1|互为相反数
|ab-2|>=0,|b-1|>=0
所以，
ab-2=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+1/2-1/3+1/3-1/4+........+1/2013-1/2014
=1-1/2014
=2013/2014

Answer 3

ab=2 b=1得到a=2 b=1 然后再拆项抵消掉