设f(x)=2/(1+x),定义fn+1(x)=f1[fn(x)],an={fn(0)-1}/{fn(0)+2},n∈N*,则数列{an}的通项公式为什么?

Question

设f(x)=2/(1+x),定义fn+1(x)=f1[fn(x)],an={fn(0)-1}/{fn(0)+2},n∈N*,则数列{an}的通项公式为什么?

Answer 1

fn+1(x)=f1[fn（x)],
fn+1(x)=2/(1+fn(x))
fn+1(x)+2=(4+2fn(x))/(1+fn(x))
fn+1(x)-1=(1-fn(x))/(1+fn(x))
[fn+1(x)+2]/[fn+1(x)-1]=(4+2fn(x))/(1-fn(x))=-2*[fn(x)+2]/[fn(x)-1]
故[fn(0)+2]/[fn(0)-1]是以 (f1(0)+2)/(f1(0)-1)＝4为首项,-2为公比的等比数列
故：(fn(0)+2)/(f(0)-1)=4*(-2)^(n-1)
an=1/4*(-1/2)^(n-1)