f1(x)=2/(1+x) 则:fn+1(0)=f1[fn(0)]=2/[1+fn(0)]所以[fn+1(0)-1]/[fn+1(0)+2]=(-1/2)*[fn(0)-1]/[fn(0)+2](化简后)即an+1=(-2)ana1=1/4a2=-1/8所以an=1/4*(-1/2)^n-1=(-1/2)^n+1对于任何正整数n均成立弯猜。埋唯型a2009=(-1/山迟2)^2010=1/(2^2010