Question

设f1(x)=2/(1+x),定义f(n+1)(x)=f1[fn(x)],an=[fn(0)-1]/[fn(0)+2]

Answer 1

(1)由条件可推出an=(-1/2)an-1（见图） 注：n-1为下标

又a1=[f1(0)-1]/[f1(0)+2]=1/4≠0,所以{an}是等比数列.

an=(1/4)•(-1/2)^(n-1)=(-1/2)^(n+1)

(2)T(2n)=(a1)+2(a2)+3(a3)+...+(2n)(a2n) ①

(-1/2)•T(2n)=(a2)+2(a3)+3(a4)+...+(2n)(a2n+1),②注：2n+1为下标

① - ②得

(3/2))•T(2n)=a1+a2+a3+...+a2n-(2n)(a2n+1)

=(1/2)[1-(-1/2)^2n]- (2n)(-1/2)^(2n+2)注：(-1/2)^(2n+2)表示-1/2的2n+2次方

=1/2-(1/2)^(2n+1) -n•(1/2)^(2n+1)

=1/2 -(n+1)•(1/2)^(2n+1)

容易证明 2^(2n+1)≥4(n+1)

所以 (n+1)•(1/2)^(2n+1) ≤1/4

(3/2))•T(2n)=1/2 -(n+1)•(1/2)^(2n+1) ≥1/2 -1/4=1/4

T(2n)≥1/6

所以9T(2n)≥3/2

而 Qn=[4(n^2)+n]/[4(n^2)+4n+1]<1

所以 9T(2n)>Qn