高一数学一小题(数列)

来源:百度知道 编辑:UC知道 时间:2024/09/24 05:23:08
两个数列{an}和{bn}满足bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)
(n∈N*).
(1)若数列{bn}是等差数列,求证:{an}也是等差数列;
(2)若数列{an}是等差数列,求证:{bn}也是等差数列。

(1)把分母乘过去
写类比式子
相减就可以了
(2)设an公差为d,则
bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)
=2(a1+2a2+3a3+…+nan)/n(n+1)
=2(a1+2(a1+d)+3(a1+2d)+…+n(a1+(n-1)d)/n(n+1)
=2{(a1+2a1+3a1+…+na1)+[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
=2{(n(n+1)a1/2)+[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
={(n(n+1)a1)+2[1*2+2*3+3*4+…(n-1)n]d}/n(n+1)
=a1+2[1*2+2*3+3*4+…+(n-1)n]d/n(n+1)
=a1+2[1+2+3+…+n-1+1^2+2^2+3^2+…+(n-1)^2]d/n(n+1)
=a1+2(n-1)n(n+1)d/3n(n+1)
=a1+(n-1)2d/3

即是bn是以a1为首数,2d/3为公差的等差数列,证毕。