Example $8 n^{2}(n \geqslant 4)$ positive numbers are arranged in several rows and columns:
\begin{tabular}{llllll}
$a_{11}$ & $a_{12}$ & $a_{13}$ & $a_{14}$ & $\ldots$ & $a_{1 n}$, \\
$a_{21}$ & $a_{22}$ & $a_{23}$ & $a_{24}$ & $\ldots$ & $a_{2 n}$, \\
$a_{31}$ & $a_{32}$ & $a_{33}$ & $a_{34}$ & $\ldots$ & $a_{3 n}$, \\
$\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ & $\ldots$ \\
$a_{n 1}$ & $a_{n 2}$ & $a_{n 3}$ & $a_{n 4}$ & $\ldots$ & $a_{n n}$,
\end{tabular}

where the numbers in each row form an arithmetic sequence, and the numbers in each column form a geometric sequence, and all the common ratios are equal. Given $a_{24}=1, a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, find $a_{11}+a_{22}+a_{33}+\cdots+a_{n n}$.
Given that an arithmetic sequence can be uniquely determined by its first term and common difference, and a geometric sequence can be uniquely determined by its first term and common ratio. If we set \(a_{11}=a\), the common difference of the first row is \(d\), and the common ratio of the first column is \(q\), it is easy to calculate that \(a_{st}=[a+(t-1)] q^{-1}\). From the given conditions, we can establish a system of equations to solve for \(n\), \(d\), and \(q\).

Solution: Let the common difference of the first row be \(d\), and the common ratio of each column be \(q\). Then the common difference of the fourth row is \(d q^{2}\). Thus, we can obtain the system of equations:
\[
\left\{\begin{array}{l}
a_{24}=\left(a_{11}+3 d\right) q=1 \\
a_{42}=\left(a_{11}+d\right) q^{3}=\frac{1}{8} \text { Solving this system of equations, we get } a_{11}=d=q= \pm \frac{1}{2} . \text { Since all } n^{2} \text { numbers } \\
a_{43}=a_{42}+d q^{3}=\frac{3}{16},
\end{array}\right.
\]

are positive, we must have \(q>0\), hence \(a_{11}=d=q=\frac{1}{2}\). Therefore, for any \(1 \leqslant k \leqslant n\), we have \(a_{k k}=a_{1 k} q^{k-1}=\left[a_{11}+(k-1)\right] q^{k-1}=\frac{k}{2^{k}}\). Thus, \(S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}\). Also, \(\frac{1}{2} S=\frac{1}{2^{2}}+\frac{2}{2^{3}}+\frac{3}{2^{4}}+\cdots+\frac{n}{2^{n+1}}\). Subtracting these two equations, we get \(\frac{1}{2} S=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}-\frac{n}{2^{n+1}}\), so \(S=2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}\).