两个数列{an}和{bn}满足bn=a1+2a2+...+nan/1+2+...+n,求证:若{bn}为等差数列,则数列{an}也是等差数列?能看懂的
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![两个数列{an}和{bn}满足bn=a1+2a2+...+nan/1+2+...+n,求证:若{bn}为等差数列,则数列{an}也是等差数列?能看懂的](/uploads/image/z/3618710-62-0.jpg?t=%E4%B8%A4%E4%B8%AA%E6%95%B0%E5%88%97%7Ban%7D%E5%92%8C%7Bbn%7D%E6%BB%A1%E8%B6%B3bn%3Da1%2B2a2%2B...%2Bnan%2F1%2B2%2B...%2Bn%2C%E6%B1%82%E8%AF%81%EF%BC%9A%E8%8B%A5%7Bbn%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%2C%E5%88%99%E6%95%B0%E5%88%97%7Ban%7D%E4%B9%9F%E6%98%AF%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%3F%E8%83%BD%E7%9C%8B%E6%87%82%E7%9A%84)
两个数列{an}和{bn}满足bn=a1+2a2+...+nan/1+2+...+n,求证:若{bn}为等差数列,则数列{an}也是等差数列?能看懂的
两个数列{an}和{bn}满足bn=a1+2a2+...+nan/1+2+...+n,求证:若{bn}为等差数列,则数列{an}也是等差数列?
能看懂的
两个数列{an}和{bn}满足bn=a1+2a2+...+nan/1+2+...+n,求证:若{bn}为等差数列,则数列{an}也是等差数列?能看懂的
设Bn公差为d
Bn=(A1+2A2+3A3+……+nAn)/(1+2+3+……+n)
=2(A1+2A2+3A3+……+nAn)/[n(n+1)][分子分母同时乘(n+2)]
=2(A1+2A2+3A3+……+nAn)(n+2)/[n(n+1)(n+2)]
B(n+1)=[A1+2A2+3A3+……+nAn+(n+1)A(n+1)]/[1+2+3+……+n+(n+1)]
=2[A1+2A2+3A3+……+nAn+(n+1)A(n+1)]/[(n+1)(n+2)][分子分母同时乘n]
=2[A1+2A2+3A3+……+nAn+(n+1)A(n+1)]n/[n(n+1)(n+2)]
由Bn+d=B(n+1),得
2n(A1+2A2+3A3+……+nAn)+4(A1+2A2+3A3+……+nAn)+n(n+1)(n+2)d=
2n(A1+2A2+3A3+……+nAn)+2n(n+1)A(n+1)
4(A1+2A2+3A3+……+nAn)+n(n+1)(n+2)d=2n(n+1)A(n+1)……1式
用n-1代换n,得
4(A1+2A2+3A3+……+(n-1)A(n-1))+(n-1)n(n+1)d=2n(n-1)An……2式
1式-2式,得
4nAn+3n(n+1)d=2n(n+1)A(n+1)-2n(n-1)An
2n(n+1)An+3n(n+1)d=2n(n+1)A(n+1)
An+1.5d=A(n+1)