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

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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设数列an的前n项和为Sn,满足an+Sn=An^2+Bn+1(

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
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

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

(1)
证:
a1+S1=A×1²+B×1+1
S1=a1 a1=3/2代入,整理,得
A+B=2 (1)
a2+S2=A×2²+B×2+1
a2+a1+a2=4A+2B+1
a1=3/2 a2=9/4代入,整理,得
4A+2B=5 (2)
联立(1),(2)解得A=1/2 B=3/2

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(1)
证:
a1+S1=A×1²+B×1+1
S1=a1 a1=3/2代入,整理,得
A+B=2 (1)
a2+S2=A×2²+B×2+1
a2+a1+a2=4A+2B+1
a1=3/2 a2=9/4代入,整理,得
4A+2B=5 (2)
联立(1),(2)解得A=1/2 B=3/2
an+Sn=(1/2)n²+(3/2)n +1
an-n +Sn=(1/2)n²+(1/2)n+1
n≥2时,
a(n-1)-(n-1)+S(n-1)=(1/2)(n-1)²+(1/2)(n-1) +1
(an -n +Sn)-[a(n-1)-(n-1)+S(n-1)]=(1/2)n²+(1/2)n +1-(1/2)(n-1)²-(1/2)(n-1)-1
整理,得
2(an -n)=a(n-1)-(n-1)
(an -n)/[a(n-1)-(n-1)]=1/2,为定值。
a1-1=3/2 -1=1/2
数列{an -n}是以1/2为首项,1/2为公比的等比数列。
(2)
由(1)得
an-n =(1/2)×(1/2)^(n-1)=1/2ⁿ
an=n +1/2ⁿ
n=1时,a1=1+1/2=3/2,n=2时,a2=2+1/4=9/4,均满足通项公式
综上,得数列{an}的通项公式为an=n +1/2ⁿ。

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