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)+··
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/11/09 10:14:58
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
···+1/(a+2013)*(b+2013)的值
|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
再问: 请告诉我理由
再答: |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
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
再问: 请告诉我理由
再答: |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
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