已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标(n+1)=a下标bn,则{bn}的通项公式

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已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标(n+1)=a下标bn,则{bn}的通项公式已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标

已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标(n+1)=a下标bn,则{bn}的通项公式
已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标(n+1)=a下标bn,则{bn}的通项公式

已知等差数列{an}满足a2=3,a5=9,若数列{bn}满足b1=3,b下标(n+1)=a下标bn,则{bn}的通项公式
a2=a1+d=3,a5=a1+4d=9,解得a1=1,d=2→an=2n-1
a(bn)=2bn-1=b(n+1)
即2bn-2=2(bn-1)=b(n+1)-1
[b(n+1)-1]/(bn-1)=2(常数）
所以{bn-1}是首项为b1-1=2,公比为2的等比数列.
bn-1=2×2^(n-1)=2^n
bn=2^n+1

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