Question

f(x)=bx+1,且g(n)=1 n=0 ,g(n)=f[g(n-1)] n∈N* .设数列{an}满足：an=g(n

Answer 1

g(n)=f[g(n-1)] f(x)=bx+1so we have g(n)=bg(n-1)+1and also g(n+1)=bg(n)+1then g(n+1)-g(n)=b(g(n)-g(n-1))that is a(n+1)=b a(n) (an=g(n)-g(n-1))so 等比数列 b is the比