数列{an}的前n项和记为sn，已知a1=1,An+1=(n+2)sn/n 1数列{sn/n}是等比数列 2sn+1=4an

1、A(n+1)=(n+2)sn/n=S(n+1)-Sn
即nS(n+1)-nSn=(n+2)Sn
nS(n+1)=(n+2)Sn+nSn
nS(n+1)=(2n+2)Sn
S(n+1)/(n+1)=2Sn/n
即S[(n+1)/(n+1)]/[Sn/n]=2
S1/1=A1=1

所以Sn/n是以2为公比1为首项的等比数列

2、由1有Sn/n是以2为公比1为首项的等比数列
所以Sn/n的通项公式是Sn/n=1*2^(n-1)
即Sn=n2^(n-1)

那么S(n+1)=(n+1)2^n,S(n-1)=(n-1)2^(n-2)
An=Sn-S(n-1)
=n2^(n-1)-(n-1)2^(n-2)
=n*2*2^(n-2)-(n-1)2^(n-2)
=[2n-(n-1)]*2^(n-2)
=(n+1)2^(n-2)
=(n+1)*2^n/2^2
=(n+1)2^n/4
=S(n+1)/4

所以有S(n+1)=4An

1、A(n+1)=S(n+1)-Sn
An+1=(n+2)sn/n 变为 S(n+1)-Sn=(n+2)Sn/n
化简整理得 nS(n+1)=2nSn+2Sn=2Sn(n+1)
so，S(n+1)/(n+1)=2*(Sn/n)可证等比

2、由1得，Sn/n=2^(n-1)
SO, Sn=n2^(n-1)
SO,S(n+1)=(n+1)2^n
又An=[(n+1)/(n-1)]*S(n-1)
=[(n+1)/(n-1)]*(n-1)*2^(n-2)
=(n+1)*2^(n-2)=(1/4)*(n-1)*2^n=(1/4)S(n+1)
SO,sn+1=4an