已知数列{an}的前N项和为Sn,a1=1.nan+1=(n+2)Snnan+1的n+1是下角标哈.求证数列{Sn/n}为等比数列求an的通项公式及前N项和Sn

na(n+1)=(n+2)Sn
n(Sn+1-Sn)=(n+2)Sn
nSn+1=2(n+1)Sn
Sn+1/(n+1)=2Sn/n
所以{Sn/n}是2为公比的等比数列
即通向为s1*2^n-1=2^n-1
an+1=(n+2)*sn/n
所以an=(n+1)2^(n-2)
因为sn/n的通向为2^n-1
所以sn=n*2^n-1

2楼答案正确 一楼错了

nan+1=(n+2)Sn,则
(n-1)an=(n+1)S(n-1)
相减,
nan+1/(n+2)-(n-1)an/(n+1)=an
即
nan+1/(n+2)=2nan/(n+1)
则,
an+1/an=2(n+2)/(n+1)
则,
(Sn/n)/(Sn/(n-1))=(an+1/(n+2))/(an/(n+1))
=(n+1)/(n+2)*2(n+2)/(n+1)
=2
则{Sn/n}为等比数列
Sn/n=(1-2^n)/(1-2)=2^n-1
则,Sn=n(2^n-1)
则S(n-1)=(n-1)(2^(n-1)-1)
相减,
an=n(2^n-1)-（n-1)(2^(n-1)-1)
得
an=(n+1)*2^(n-1)+1

将An+1=Sn+1-Sn代入nAn+1=(n+2)Sn可得
n（Sn+1-Sn）=(n+2)Sn
即nSn+1=(2n+2)Sn
即Sn+1/n+1=2*Sn/n
则Sn/n是以S1/1=A1/1=1为首项，2为公比的等比数列
则Sn/n=2^(n-1),Sn=n*2^(n-1);
当n>=2时,An=Sn-Sn-1=n*2^(n-1)-(n-1)*2^(n-2)=(n+1)*2^(n-2)
A1=1也满足上式，则An=(n+1)*2^(n-2)，（n=1，2，3，。。。）
上面已求出Sn=n*2^(n-1)