10. (25 points) For any positive integers $m, n$, define the function $f(m, n)$ as follows:
(i) $f(1,1)=1$;
(ii) $f(m+1, n)=f(m, n)+2(m+n)$;
(iii) $f(m, n+1)=f(m, n)+2(m+n-1)$.
(1) Find the analytical expression for $f(m, n)$;
(2) Let $a_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}\left(n \in \mathbf{Z}_{+}\right), S_{n}$ be the sum of the first $n$ terms of the sequence $\left\{a_{n}\right\}$, prove: $S_{n}<6$.
10. (1) From conditions (i) and (ii), we have
$$
\begin{array}{l}
f(m, 1)-f(1,1) \\
=\sum_{k=1}^{m-1}(f(k+1,1)-f(k, 1)) \\
=2 \sum_{k=1}^{m-1}(k+1) \\
\Rightarrow f(m, 1)=m^{2}+m-1 \\
\Rightarrow f(m, n)-f(m, 1) \\
\quad=\sum_{k=1}^{n-1}(f(m, k+1)-f(m, k)) \\
\quad=2 \sum_{k=1}^{n-1}(m+k-1) \\
\Rightarrow f(m, n)=(m+n)^{2}-m-3 n+1 .
\end{array}
$$
(2) From (1), we get $f(n, n)=(2 n-1)^{2}$.

Then $a_{n}=\frac{\sqrt{f(n, n)}}{2^{n-1}}=(2 n-1)\left(\frac{1}{2}\right)^{n-1}$.
Thus, $S_{n}=\sum_{k=1}^{n}(2 k-1)\left(\frac{1}{2}\right)^{k-1}$,
$$
\frac{1}{2} S_{n}=\sum_{k=1}^{n}(2 k-1)\left(\frac{1}{2}\right)^{k} \text {. }
$$

Subtracting the two equations, we get
$$
\begin{array}{l}
\frac{1}{2} S_{n}=1+2 \sum_{k=1}^{n-1}\left(\frac{1}{2}\right)^{k}-(2 n-1)\left(\frac{1}{2}\right)^{n} \\
=3-\frac{1}{2^{n-2}}-\frac{2 n-1}{2^{n}} .
\end{array}
$$

Therefore, $S_{n}=6-\frac{2 n+3}{2^{n-1}}<6$.