{an}首项为 1,前n项和为Sn,任意正整数n 有n an Sn成等差数列 1证数列{Sn+n+2}成等比数列2求{an}通项公式

Sn+n=2*An
则Sn + A(n+1) + n + 1=2 * A(n+1),,,A1=1;
∴A(n+1) - 2*An=1
∴ (A(n+1) + 1)=2*(An + 1)
∴ A1 + 1=2, An=2^n - 1

设数列{An+1}的和为Tn;
∴ Tn=2(1-2^n)/(1-2)= 2*2^n -2 =Sn + n
∴ Sn= 2^(n+1) - 2 - n
∴ Sn + n + 2 = 2^(n+1)
∴ {Sn+n+2}成等比数列,公比为2;

（1）n、an、Sn成等差数列,则2an=n+Sn,用an=Sn-S(n-1)代入,整理得Sn=2S(n-1)+n,等号两边同时加上n+2,得Sn+n+2=2[S(n-1)+n-1+2],所以数列{Sn+n+2}成等比数列.
（2）由（1）可知{Sn+n+2}的通项公式Sn+n+2=4*2^(n-1)=2^(n+1),然后由递推得S(n-1)+n-1+2=2^n,两式相减得an+1=2^n,所以{an}的的通项公式为an=2^n-1.