设f1(x)=2/1+x,定义fn+1(x)=f1[fn(x)],an=fn(0)−1fn(0)+2,其中n∈N*,则数列{an}的通项_.
问题描述:
设f1(x)=
,定义fn+1(x)=f1[fn(x)],an=2 1+x
,其中n∈N*,则数列{an}的通项______.
fn(0)−1
fn(0)+2
答
(1)∵f1(0)=2,a1=2−12+2=14,fn+1(0)=f1[fn(0)]=21+fn(0),∴an+1=fn+1(0)−1fn+1(0)+2=21+fn(0)−121+fn(0)+2=1−fn(0)4+2fn(0)=-12•fn(0)−1fn(0)+2=-12an,∴q=an+1an=-12,∴数列{an}是首项为14,公...