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已知数列{an}的前n项和为Sn,且Sn=n-5an-85,n∈N*.
(1)证明:{an-1}是等比数列;
(2)求数列{Sn}的通项公式.请指出n为何值时,Sn取得最小值,并说明理由.
答案:
(1)∵Sn=n-5an-85,n∈N*.
∴Sn+1=n+1-5an+1-85,n∈N*.
两式作差得an+1=1-5an+1+5an,即6(an+1-1)=5(an-1),即(an+1-1)=5/6(an-1),n∈N*.
故{an-1}是等比数列
(2)由(1)Sn+1=n+1-5an+1-85,n∈N*.得Sn+1=n+1-5(Sn+1-Sn)-85,n∈N*.
得6Sn+1=n+5Sn-84,即6[Sn+1-(n+1)]=5(Sn-n)-90,
即Sn+1-(n+1)=5/6(Sn-n)-15
整理得Sn+1-(n+1)+90=5/6(Sn-n+90)
故{Sn-n+90}是一个等比数列,其公比为5/6
,由于a1=1-5a1-85,得a1=-14
故{Sn-n+90}的首项为-14-1+90=75
故Sn-n+90=75×(5/6)n−1,即Sn=n-90+75×(5/6)n−1,
由于Sn+1-Sn=1-75/6×(5/6)n−1,令Sn+1-Sn>0,对n赋值验证知n>15时成立,即Sn其最小值是S15
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