设数列{an}的前n项和为Sn,已知ban-2n=(b-1)Sn
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设数列{an}的前n项和为Sn,已知ban-2n=(b-1)Sn
(1)证明:当b=2时,{a
(1)证明:当b=2时,{a
(1)证明:由题意知a1=2,且ban-2n=(b-1)Sn,
ban+1-2n+1=(b-1)Sn+1
两式相减得b(an+1-an)-2n=(b-1)an+1
即an+1=ban+2n①
当b=2时,由①知an+1=2an+2n
于是an+1-(n+1)•2n=2an+2n-(n+1)•2n=2(an-n•2n-1)
又a1-1•20=1≠0,所以{an-n•2n-1}是首项为1,公比为2的等比数列;
(2)当b=2时,由(1)知an-n•2n-1=2n-1,即an=(n+1)2n-1,
当b≠2时,由①得an+1-
1
2−b•2n+1=ban+2n-
1
2−b•2n+1=ban-
b
2−b•2n=b(an-
1
2−b•2n)
因此an+1-
1
2−b•2n+1=b(an-
1
2−b•2n)=
2(1−b)
2−b•bn,
即an+1=
1
2−b•2n+1+
2(1−b)
2−b•bn,
所以an=
1
2−b•2n+
2(1−b)
2−b•bn-1
ban+1-2n+1=(b-1)Sn+1
两式相减得b(an+1-an)-2n=(b-1)an+1
即an+1=ban+2n①
当b=2时,由①知an+1=2an+2n
于是an+1-(n+1)•2n=2an+2n-(n+1)•2n=2(an-n•2n-1)
又a1-1•20=1≠0,所以{an-n•2n-1}是首项为1,公比为2的等比数列;
(2)当b=2时,由(1)知an-n•2n-1=2n-1,即an=(n+1)2n-1,
当b≠2时,由①得an+1-
1
2−b•2n+1=ban+2n-
1
2−b•2n+1=ban-
b
2−b•2n=b(an-
1
2−b•2n)
因此an+1-
1
2−b•2n+1=b(an-
1
2−b•2n)=
2(1−b)
2−b•bn,
即an+1=
1
2−b•2n+1+
2(1−b)
2−b•bn,
所以an=
1
2−b•2n+
2(1−b)
2−b•bn-1
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