已知数列{an}和{bn}满足:bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)求当{an}是等差数列
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已知数列{an}和{bn}满足:bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)求当{an}是等差数列的时候证明{bn}是等差数
![已知数列{an}和{bn}满足:bn=(a1+2a2+3a3+…+nan)/(1+2+3+…+n)求当{an}是等差数列](/uploads/image/z/16117848-0-8.jpg?t=%E5%B7%B2%E7%9F%A5%E6%95%B0%E5%88%97%7Ban%7D%E5%92%8C%7Bbn%7D%E6%BB%A1%E8%B6%B3%3Abn%3D%28a1%2B2a2%2B3a3%2B%E2%80%A6%2Bnan%29%2F%281%2B2%2B3%2B%E2%80%A6%2Bn%29%E6%B1%82%E5%BD%93%7Ban%7D%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97)
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.
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.
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