设数列{an}的前n项和为Sn,已知a1=a,an+1=Sn+3^n,n∈N+.设bn=Sn+3n,求数列{bn}的通项

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设数列{an}的前n项和为Sn,已知a1=a,an+1=Sn+3^n,n∈N+.设bn=Sn+3n,求数列{bn}的通项公式

由an+1=Sn+3n得：Sn+1-Sn =Sn+3n,即Sn+1=2Sn+3n．所以Sn+1-3n+1=2Sn+3n-3n+1．
整理得：Sn+1-3n+1=2(Sn-3n),这就是说,数列{ Sn-3n }是以a-3为首项,以2为公比的等比数列,故Sn-3n =(a-3) ∙2 n-1,即Sn=(a-3) ∙2 n-1+3n,所以bn=Sn-3n=(a-3) ∙2 n-1．

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