已知无穷数列{an}的前n项和为Sn,且满足Sn=Aan^2+Ban+C,其中A,B,C是常数（1）若A=0,B=3,C=-2,求数列{an}的通项公式（2）若A=1,B=1/2,C=1/16,且an>0,求数列{an}的前n项和Sn;(3) 试探究A,B,C满足什么条件时,数列{an}

已知无穷数列{an}的前n项和为Sn,且满足Sn=Aan^2+Ban+C,其中A,B,C是常数（1）若A=0,B=3,C=-2,求数列{an}的通项公式（2）若A=1,B=1/2,C=1/16,且an>0,求数列{an}的前n项和Sn;(3) 试探究A,B,C满足什么条件时,数列{an}
已知无穷数列{an}的前n项和为Sn,且满足Sn=Aan^2+Ban+C,其中A,B,C是常数
（1）若A=0,B=3,C=-2,求数列{an}的通项公式
（2）若A=1,B=1/2,C=1/16,且an>0,求数列{an}的前n项和Sn;
(3) 试探究A,B,C满足什么条件时,数列{an}是公比不为-1的等比数列.

已知无穷数列{an}的前n项和为Sn,且满足Sn=Aan^2+Ban+C,其中A,B,C是常数（1）若A=0,B=3,C=-2,求数列{an}的通项公式（2）若A=1,B=1/2,C=1/16,且an>0,求数列{an}的前n项和Sn;(3) 试探究A,B,C满足什么条件时,数列{an}
s(n) = A[a(n)]^2 + Ba(n) + C.
(1) s(n) = 3a(n) - 2,
a(1) = s(1) = 3a(1) - 2 ,a(1) = 1.
s(n+1) = 3a(n+1) -2.
a(n+1) = s(n+1)-s(n) = 3a(n+1) - 3a(n),
a(n+1) = (3/2)a(n),
{a(n)}是首项为a(1)=1,公比为3/2的等比数列.
a(n) = (3/2)^(n-1).
(2) s(n) = [a(n)]^2 + a(n)/2 + 1/16,a(n)>0.
a(1) = s(1) = [a(1)]^2 + a(1)/2 + 1/16,0 = [a(1)]^2 - a(1)/2 + (1/4)^2 = [a(1) - 1/4]^2,
a(1) = s(1) = 1/4.
s(n+1) = [a(n+1)]^2 + a(n+1)/2 + 1/16,
a(n+1) = s(n+1)-s(n) = [a(n+1)]^2 + a(n+1)/2 - [a(n)]^2 - a(n)/2,
0 = [a(n+1)]^2 - [a(n)]^2 - a(n+1)/2 - a(n)/2 = [a(n+1)+a(n)][a(n+1)-a(n)] - [a(n+1)+a(n)]/2
= [a(n+1)+a(n)][a(n+1)-a(n)-1/2],
因a(n)>0,所以,a(n+1)+a(n)>0.
0 = a(n+1)-a(n)-1/2,
a(n+1) = a(n) + 1/2.
{a(n)}是首项为a(1)=1/4,公差为1/2的等差数列.
a(n) = 1/4 + (n-1)/2 = (2n-1)/4.
s(n) = [a(n)]^2 + a(n)/2 + 1/16 = (2n-1)^2/16 + (2n-1)/8 + 1/16 = [(2n-1)^2 + 2(2n-1) + 1]/16
= (2n-1+1)^2/16
= n^2/4
(3) a(n) = aq^(n-1),q不为-1.
s(n) = A[a(n)]^2 + Ba(n) + C = Aa^2q^(2n-2) + Baq^(n-1) + C,
a = a(1) = s(1) = Aa^2 + Ba + C,
0 = Aa^2 + (B-1)a + C.必须有实数解.
因此,必须有,Delta = (B-1)^2 - 4AC >= 0.
s(n+1) = Aa^2q^(2n) + Baq^n + C,
aq^n = a(n+1) = s(n+1) - s(n) = Aa^2q^(2n) + Baq^n - Aa^2q^(2n-2) - Baq^(n-1),
q = Aaq^(n+1) + Bq - Aaq^(n-1) - B,
0 = Aaq^(n+1) - Aaq^(n-1) + (B-1)q - B
0 = Aaq^(n-1)(q^2 - 1) + (B-1)q - B.对任意正整数n都成立,只能,
(3.1)q=1,此时a(n) = a,s(n) = na = Aa^2 + Ba + C,
na = Aa^2 + Ba + C对任意正整数n成立,只能,a=0,与a(n)=aq^(n-1)是等比数列矛盾.
因此,q不为1.
(3.2)q不为1时,只能A = 0.此时,
0 = Aaq^(n-1)(q^2 - 1) + (B-1)q-B = (B-1)q - B.
若B=1,则0 = -B,0 = B与B=1矛盾.因此,B不为1.
q = B/(B-1),
0 = Aa^2 + (B-1)a + C = (B-1)a + C,
a = -C/(B-1).
a(n) = -C/(B-1)[B/(B-1)]^(n-1),
s(n) = -C/(B-1){[B/(B-1)]^n - 1}/[B/(B-1) - 1] = -C{[B/(B-1)]^n - 1 }
= -CB/(B-1)[B/(B-1)]^(n-1) + C = Ba(n) + C,满足题意.
====验证=====
A=0,B不为1时,
s(n) = Ba(n) + C,
a(1) = s(1) = Ba(1) + C,a(1) = -C/(B-1).
s(n+1)= Ba(n+1) + C,
a(n+1) = s(n+1)-s(n) = Ba(n+1)-Ba(n),
a(n+1) = [B/(B-1)]a(n),
{a(n)}是首项为a(1)=-C/(B-1),公比为B/(B-1)的等比数列.
====验证完毕====
综合,有,
A=0,B不为1时,{a(n)}是公比不为-1的等比数列.

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