Example 3 (1990 National High School League Question) $n^{2}(n \geqslant 4)$ positive numbers are arranged in $n$ rows and $n$ columns:
$$
\begin{array}{l}
a_{11} a_{12} a_{13} a_{14} \cdots a_{1 n} \\
a_{21} a_{22} a_{23} a_{24} \cdots a_{2 n} \\
a_{31} a_{32} a_{33} a_{34} \cdots a_{3 n} \\
a_{41} a_{42} a_{43} a_{44} \cdots a_{4 n} \\
\cdots \\
a_{n 1} a_{n 2} a_{n 3} a_{n 4} \cdots a_{m}
\end{array}
$$

where the numbers in each row form an arithmetic sequence, and the numbers in each column form a geometric sequence, with all common ratios being equal. Given that $a_{24}=1$, $a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, find $a_{11}+a_{22}+\cdots+a_{m}$.
Let the common difference of the first row arithmetic sequence be $d$, and the common ratio of each column geometric sequence be $q$. Then, the common difference of the $k$-th row arithmetic sequence is $d \cdot q^{k-1}$. Therefore,
$$
\left\{\begin{array}{l}
a_{24}=a_{14} q=\left(a_{11}+3 d\right) q=1, \\
a_{42}=a_{12} q^{3}=\left(a_{11}+d\right) q^{3}=\frac{1}{8}, \\
a_{43}=a_{13} q^{3}=\left(a_{11}+2 d\right) q^{3}=\frac{3}{16} .
\end{array}\right.
$$

Solving the system of equations, we get $a_{11}=d=q=\frac{1}{2}$. Hence,
$$
a_{b k}=a_{1 k} q^{k-1}=\left[a_{11}+(k-1) d\right] q^{k-1}=\left[\frac{1}{2}+(k-1) \frac{1}{2}\right]\left(\frac{1}{2}\right)^{k-1}=\frac{k}{2^{k}} .
$$

Let $S_{n}=a_{11}+a_{22}+\cdots+a_{m}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\cdots+\frac{n}{2^{n}}$, then
$$
\begin{aligned}
S_{n} & =2 S_{n}-S_{n}=\left(1+\frac{2}{2}+\frac{3}{2^{2}}+\cdots+\frac{n}{2^{n-1}}\right)-\left(\frac{1}{2}+\frac{3}{2^{2}}+\cdots+\frac{n}{2^{n}}\right) \\
& =1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n-1}}-\frac{n}{2^{n}}=\frac{1-\frac{1}{2^{n}}}{1-\frac{1}{2}}-\frac{n}{2^{n}} \\
& =2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}} .
\end{aligned}
$$