f1(x)=2/(1+x),
f(n+1)(x)=f1[fn(x)]=2/[1+fn(x)]
f(n+1)(x)=2/[1+fn(x)]
f(n+1)(x)-1=2/[1+fn(x)]-1=[1-fn(x)]/[1+fn(x)]
f(n+1)(x)+2=2/[1+fn(x)]+2=2[2+fn(x)]/[1+fn(x)]
两式相除：
2[f(n+1)(x)-1]/[f(n+1)(x)+2]=[1-fn(x)]/[2+fn(x)]=-[fn(x)-1]/[2+fn(x)]
当x=0时,2[f(n+1)(0)-1]/[f(n+1)(0)+2]=-[fn(0)-1]/[2+fn(0)]
2a(n+1)=-an
an=[(-1/2)^(n-1)]a1
又a1=[f1(0)-1]/[f1(0)+2]
=1/4
an=[(-1/2)^(n+1)]
a2007=[(-1/2)^2008=1/2^2008