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+1 B.bn=2n+1 C.bn=3n+2 D.bn=2n+2
问题描述:
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+1
B. bn=2n+1
C. bn=3n+2
D. bn=2n+2
答
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