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
<p>问题：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<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">刘筱秀的回答：<div class="content-b">网友采纳 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)∙2n-1,即Sn=(a-3)ͨ...
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