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)