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

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 $a_{24}=1, a_{42}=\frac{1}{8}, a_{43}=\frac{3}{16}$, find $a_{11}+a_{22}+a_{23}+\cdots+a_{n n}$.
7. Let the common difference of the first row sequence be $d$, and the common ratio of each column sequence be $q$. Then the common difference of the 4th row sequence is $d q^{3}$. Thus, from $a_{24}=\left(a_{11}+3 d\right) q=1$, $a_{42}=\left(a_{11}+d\right) q^{3}=\frac{1}{8}$, and $a_{43}=\frac{1}{8}+d q^{3}=\frac{3}{16}$, we solve to get $a_{11}=d=q= \pm \frac{1}{2}$. Since the $n^{2}$ numbers are positive, we have $a_{11}=d=q=\frac{1}{2}$. For any $1 \leqslant k \leqslant n$, we have $a_{k k}=a_{1 k} q^{k-1}=k \frac{1}{2^{k}}$. Therefore, $S=a_{11}+a_{22}+\cdots+a_{n n}=\frac{1}{2}+2 \frac{1}{2^{2}}+\cdots+n \frac{1}{2^{n}}$, $\frac{1}{2} S=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\cdots+n \frac{1}{2^{n+1}}$. Subtracting these two equations gives $S=2-\frac{1}{2^{n-1}}-\frac{n}{2^{n}}$.