你題目寫錯(cuò)了,{Bn}的表達(dá)式應(yīng)該是
Bn=（A1+2A2+3A3+……+nAn)/(1+2+3+……+n)
那啥,第n+1項(xiàng)我直接用B(n+1)來表示,你應(yīng)該能看懂
設(shè)Bn公差為d
Bn=（A1+2A2+3A3+……+nAn)/(1+2+3+……+n)
=2(A1+2A2+3A3+……+nAn)/(n(n+1))
=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))
=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代換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)
得證……