若数列{an}满足a1=1,an+1=2an+3n,则数列的项a5=_______.

问题描述:

若数列{an}满足a1=1,an+1=2an+3n,则数列的项a5=_______.

∵数列{an}满足a1=1,an+1=2an+3n,
∴a2=2a1+3=5,同理可得a3=16,a4=41,a5=94.
故答案为:94.
答案解析:数列{an}满足a1=1,an+1=2an+3n,可得a2=2a1+3=5,同理可得a3,a4,a5
考试点:数列的函数特性.


知识点:本题考查了递推式的应用,属于基础题.