Example 2 Given the sequence $a_{k}=2^{k}(1 \leqslant k \leqslant n)$. Then the sum of all possible products $a_{i} a_{j}(1 \leqslant i \leqslant j \leqslant n)$ is
(1997, Shanghai High School Mathematics Competition)
Solution: Given $a_{i} a_{j}=2^{i+j}(1 \leqslant i \leqslant j \leqslant n)$, the table is as follows: $\square$
$$
\begin{array}{ccccc}
2^{1+1} & 2^{1+2} & 2^{1+3} & \cdots & 2^{1+n} \\
& 2^{2+2} & 2^{2+3} & \cdots & 2^{2+n} \\
& & 2^{3+3} & \cdots & 2^{3+n} \\
& & & \cdots & \cdots \\
& & & & 2^{n+n}
\end{array}
$$

By filling the lower triangular part symmetrically about the main diagonal, we get the square table:
\begin{tabular}{ccccc}
$2^{1+1}$ & $2^{1+2}$ & $2^{1+3}$ & $\cdots$ & $2^{1+n}$ \\
$2^{2+1}$ & $2^{2+2}$ & $2^{2+3}$ & $\cdots$ & $2^{2+n}$ \\
$2^{3+1}$ & $2^{3+2}$ & $2^{3+3}$ & $\cdots$ & $2^{3+n}$ \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ \\
$2^{n+1}$ & $2^{n+2}$ & $2^{n+3}$ & $\cdots$ & $2^{n+n}$
\end{tabular}

Let the desired sum be $S$, and the sum of the first row be $S_{1}$. Then we have
$$
\begin{aligned}
2 S= & S_{1}\left(1+2+2^{2}+\cdots+2^{n}\right) \\
& +\left(2^{2}+2^{4}+2^{6}+\cdots+2^{2 n}\right) \\
= & 2^{2}\left(2^{n}-1\right)^{2}+\frac{2^{2}}{3}\left(2^{2 n}-1\right) \\
= & \frac{4}{3}\left(2^{n}-1\right)\left(2^{n+2}-2\right) .
\end{aligned}
$$

Therefore, $S=\frac{4}{3}\left(2^{n}-1\right)\left(2^{n+1}-1\right)$ is the desired result.