f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=

f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=

f1(x)=2/(x+1),而fn+1=f1[fn(x)],设an=[fn(2)-1]/[fn(2)+2],则a99=
fn+1=2/[fn(x)+1] f1(2)=2/3 a1=-1/8,f2(2)=6/5,a2=1/16,f3(2)=10/11,a3=-1/32,所以通式an=(-1)^n*1/[2^(n+2)];a99=-1/2^101;(说明：2^n,2的n次方