10. (20 points) Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=2, a_{1} a_{2} \cdots a_{n-1}=a_{n}(n \geqslant 2), \\
T_{n}=\sum_{k=1}^{n}\left[k(k-1) \log _{a_{k}} 2\right]=16-f(n) .
\end{array}
$$

Find the expression for $f(n)$.
10. From the given, we know $a_{n} \neq 0$, and $a_{n}=\frac{a_{n+1}}{a_{n}}$.

Thus, $a_{n}=a_{n-1}^{2}=a_{n-2}^{2^{2}}=\cdots=a_{2}^{2^{n-2}}=2^{2^{n-2}}(n \geqslant 2)$.
Therefore, $a_{n}=\left\{\begin{array}{ll}2, & n=1 ; \\ 2^{2^{n-2}}, & n \geqslant 2 .\end{array}\right.$
Hence, $T_{n}=\sum_{k=1}^{n} k(k-1) \log _{a_{k}} 2$
$=\sum_{k=1}^{n} \frac{k(k-1)}{2^{k-2}}=4 \sum_{k=1}^{n}\left(\frac{k^{2}}{2^{k}}-\frac{k}{2^{k}}\right)$.
Let $S_{n}=\sum_{k=1}^{n} \frac{k}{2^{k}}$.
Then $S_{n}=2 S_{n}-S_{n}$
$=1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}-\frac{n}{2^{n}}=2-\frac{n+2}{2^{n}}$.
Let $Q_{n}=\sum_{k=1}^{n} \frac{k^{2}}{2^{k}}$.
Then $Q_{n}=2 Q_{n}-Q_{n}=\sum_{k=1}^{n} \frac{k^{2}-(k-1)^{2}}{2^{k-1}}-\frac{n^{2}}{2^{n}}$
$$
\begin{array}{l}
=\sum_{k=1}^{n} \frac{2 k-1}{2^{k-1}}-\frac{n^{2}}{2^{n}} \\
=4 \sum_{k=1}^{n} \frac{k}{2^{k}}-2 \sum_{k=1}^{n} \frac{1}{2^{k}}-\frac{n^{2}}{2^{n}} \\
=4\left(2-\frac{n+2}{2^{n}}\right)-2\left(1-\frac{1}{2^{n}}\right)-\frac{n^{2}}{2^{n}} \\
=6-\frac{n^{2}+4 n+6}{2^{n}} .
\end{array}
$$

Therefore, $T_{n}=4\left(Q_{n}-S_{n}\right)$
$$
\begin{array}{l}
=4\left(6-\frac{n^{2}+4 n+6}{2^{n}}-2+\frac{n+2}{2^{n}}\right) \\
=16-\frac{n^{2}+3 n+4}{2^{n-2}} .
\end{array}
$$

Thus, $f(n)=\frac{n^{2}+3 n+4}{2^{n-2}}$.