20 Let $f_{1}(x)=\frac{2}{1+x}, f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$, $a_{n}=\frac{f_{n}(0)-1}{f_{n}(0)+2}$, where $n \in \mathbf{N}^{*}$.
(1) Find the general term formula for the sequence $\left\{a_{n}\right\}$;
(2) If
$$
\begin{aligned}
T_{2 n} & =a_{1}+2 a_{2}+3 a_{3}+\cdots+2 n a_{2 n}, \\
Q_{n} & =\frac{4 n^{2}+n}{4 n^{2}+4 n+1},
\end{aligned}
$$

where $n \in \mathbf{N}^{*}$, compare the size of $9 T_{2 n}$ and $Q_{n}$, and provide the reasoning.
20 (1) From the problem, we know $f_{1}(0)=2, a_{1}=\frac{1}{4}$,
$$
\begin{aligned}
f_{n+1}(0) & =f_{1}\left[f_{n}(0)\right]=\frac{2}{1+f_{n}(0)} \\
a_{n+1} & =\frac{f_{n+1}(0)-1}{f_{n+1}(0)+2} \\
& =\frac{1-f_{n}(0)}{4+2 f_{n}(0)} \\
& =-\frac{1}{2} \cdot \frac{f_{n}(0)-1}{f_{n}(0)+2} \\
& =-\frac{1}{2} a_{n}
\end{aligned}
$$

Thus, $\left\{a_{n}\right\}$ is a geometric sequence with the first term $\frac{1}{4}$ and common ratio $-\frac{1}{2}$, so $a_{n}=\frac{1}{4} \cdot\left(-\frac{1}{2}\right)^{n-1}$.
(2) From the problem, we know
$$
\begin{aligned}
T_{2 n} & =a_{1}+2 a_{2}+3 a_{3}+\cdots+2 n a_{2 n}, \\
-\frac{1}{2} T_{2 n} & =a_{2}+2 a_{3}+\cdots+(2 n-1) a_{2 n}+2 n a_{2 n+1} .
\end{aligned}
$$

Subtracting the two equations, we get
$$
\begin{aligned}
\frac{3}{2} T_{2 n}= & \left(a_{1}+a_{2}+\cdots+a_{2 n}\right)-2 n \cdot \frac{1}{4} \cdot\left(-\frac{1}{2}\right)^{2 n} \\
& =\frac{1}{6}-\frac{1}{6} \cdot\left(-\frac{1}{2}\right)^{2 n}+\frac{n}{4}\left(-\frac{1}{2}\right)^{2 n-1}
\end{aligned}
$$

Therefore,
$$
T_{2 n}=\frac{1}{9}\left(1-\frac{3 n+1}{2^{2 n}}\right)
$$

And
$$
\begin{array}{l}
9 T_{2 n}=1-\frac{3 n+1}{2^{2 n}} \\
Q_{n}=\frac{n(4 n+1)}{(2 n+1)^{2}}
\end{array}
$$

The original proposition is transformed into comparing
$$
1-\frac{4 n^{2}+n}{4 n^{2}+4 n+1}=\frac{3 n+1}{(2 n+1)^{2}}
$$

with $\frac{3 n+1}{2^{2 n}}$, i.e., comparing $(2 n+1)^{2}$ with $2^{2 n}$.
When $n=1$, $2^{2 n}=4\left(1+n+\frac{n(n-1)}{2}\right)^{2} \\
& >(1+n+n)^{2} \\
& =(2 n+1)^{2}
\end{aligned}
$$

Thus, when $n=1$ or 2, $9 T_{2 n}Q_{n}$.