8-11 $n^{2}(n \geqslant 4)$ positive numbers are arranged in $n$ rows and $n$ columns:
$$\begin{array}{l}
a_{11} a_{12} a_{13} a_{14} \cdots a_{1 n} \\
a_{21} a_{22} a_{23} a_{24} \cdots a_{2 n} \\
a_{31} a_{32} a_{33} a_{34} \cdots a_{3 n} \\
a_{41} a_{42} a_{43} a_{44} \cdots a_{4 n} \\
\cdots \cdots \\
a_{n 1} a_{n 2} a_{n 3} a_{n 4} \cdots a_{n n}
\end{array}$$

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
$$a_{24}=1, a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$$
$$\text { Find } \quad a_{11}+a_{22}+\cdots+a_{m m} \text {. }$$
[Solution] Since each row of numbers forms an arithmetic sequence, and $a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, the common difference of the fourth row is $\frac{1}{16}$, and thus
$$a_{4 k}=\frac{k}{16}, k=1,2,3,4, \cdots, n$$

Furthermore, since each column of numbers forms a geometric sequence with the same common ratio, using $a_{24}=1, a_{44}=\frac{1}{4}$, we find that the common ratio is $\frac{1}{2}$. Therefore, we have $a_{k k}=\left(\frac{1}{2}\right)^{k-4} \frac{k}{16}, k=1,2,3,4, \cdots, n$. This implies $\sum_{k=1}^{n} a_{k k}=\sum_{k=1}^{n} k\left(\frac{1}{2}\right)^{k}$.
Let $S=\sum_{k=1}^{n} a_{k k}$, then
$$\frac{S}{2}=\sum_{k=1}^{n} k\left(\frac{1}{2}\right)^{k+1}=\sum_{k=2}^{n}(k-1)\left(\frac{1}{2}\right)^{k}+n\left(\frac{1}{2}\right)^{n+1}$$

Thus, $\frac{S}{2}=S-\frac{S}{2}=\frac{1}{2}+\sum_{k=2}^{n}\left(\frac{1}{2}\right)^{k}-n\left(\frac{1}{2}\right)^{n+1}$
$$=\frac{1}{2}+\frac{1}{2}-\frac{1}{2^{n}}-\frac{n}{2^{n+1}}=1-\frac{n+2}{2^{n+1}}$$

Therefore, $S=2-\frac{n+2}{2^{n}}$.