设数列an的前n项和为Sn,满足an+Sn=An^2+Bn+1（A不等于0）,a1=3/2,a2=9/4,求证an-n为

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设数列an的前n项和为Sn,满足an+Sn=An^2+Bn+1（A不等于0）,a1=3/2,a2=9/4,求证an-n为等比数列并求an

解析：当n=1时,a1+S1=A+B+1；即A+B=2；
当n=2时,a2+S2=2a2+a1=2*9/4+3/2=6=4A+2B+1,即4A+2B=5
联立以上两式,可得A=1/2,B=3/2.
由题意：an+Sn=An^2+Bn+1 ①
则a(n+1)+S(n+1)=A(n+1)^2+B(n+1)+1 ②
②-①,可得：
a(n+1)-an+a(n+1)=A(2n+1)+B=n+2
2a(n+1)-an=n+2
2a(n+1)-2(n+1) = an -n
2[a(n+1)-(n+1)]=an -n
即：[a(n+1)-(n+1)]/(an -n) =1/2
由等比数例定义可知,{an-n}为以a1-1=1/2,q=1/2的等比数列.
an-n = (1/2)^n
an = n+ (1/2)^n

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