Question

已知|ab-2|与|b-1|互为相反数,求代数式:1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+···

···+1/(a+2013)*(b+2013)的值

Answer 1

|ab-2|与|b-1|互为相反数
得ab-2=0，b-1=0
即a=2，b=1
1/ab+1/(a+1)*(b+1)+1/(a+2)*(b+2)+···
=1/1*2+1/2*3+1/3*4+......+1/n*n+1)
=(1-1/2)+(1/2-1/3）+（1/3-1/4）+...（1/n-1/(n+1))
=1-1/(n+1)
=n/(n+1)

Answer 2

|ab-2|=|b-1|=0
a=2，b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2013)(b+2013)
=1-1/2015=2014/2015

追问

请告诉我理由

追答

|ab-2|=|b-1|=0
a=2，b=1
1/ab+1/(a+1)(b+1)+1/(a+2)(b+2)+…+1/(a+2013)(b+2013)
= 1/1x2+1/2x3 +1/3x4 +…+1/2014x2015
= 1- 1/2 + 1/2-1/3 +1/3-1/4 + ....+ 1/2014-2015
=1-1/2015
=2014/2015