已知函数f(x)=x+3,又数列{an}中,a1=f(-1),a(n+1)=f(an),n属于N*,求数列{an}的通项

a(n+1)=f(an),a(n+1)=an+3
a1=f(-1)=2
an=3n-1
bn=a(n)+2n=5n-1
Sn=5(1+2+3+...+n)-n=5/2*n^2+3/2*n