已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,bn+1=abn,则{bn}的通项公式为()A.bn=3n+1B.bn=2n+1C.bn=3n+2D.bn=2n+2-数学

题目简介

已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,bn+1=abn,则{bn}的通项公式为()A.bn=3n+1B.bn=2n+1C.bn=3n+2D.bn=2n+2-数学

题目详情

已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3, bn+1=abn,则{bn}的通项公式为(  )
A.bn=3n+1B.bn=2n+1C.bn=3n+2D.bn=2n+2
题型:解答题难度:中档来源:不详

答案

由已知,等差数列{an},d=2,则{an}通项公式an=2n-1,bn+1=2bn-1 
两边同减去1,得b n+1-1=2(bn-1 )
∴数列{bn-1}是以2为首项,以2为公比的等比数列,
bn-1=2×2 n-1=2n,
∴bn=2n+1
故选B

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