Example 4 Find the sum $S_{n}=n \cdot 1+(n-1) \cdot 3+(n$ $-2) \cdot 5+\cdots+1 \cdot(2 n-1)$.
Solution: Each term in $S_{n}$ can be viewed as the sum of 1 $n$, 3 $n-1$s, ..., and $2n-1$ 1s. Thus, we can construct the following triangular number table (for convenience, only the case for $n=5$ is shown):
$$
\begin{array}{lllllllll}
1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\
& 2 & 2 & 2 & 2 & 2 & 2 & 2 & \\
& & 3 & 3 & 3 & 3 & 3 & & \\
& & & 4 & 4 & 4 & & & \\
& & & & 5 & & & & \\
& & & & &
\end{array}
$$

Transform this table into a new triangular number table:

It is easy to see that the sum of each row in the new table is a perfect square, i.e., 1
$$
\begin{array}{l}
=1^{2}, 1+2+1=2^{2}, \cdots, 1+2+\cdots+(n-1)+ \\
n+(n-1)+\cdots+2+1=n^{2} .
\end{array}
$$

Therefore, $S_{n}=\sum_{k=1}^{n} k^{2}=\frac{1}{6} n(n+1)(2 n+1)$.
Comment: This solution demonstrates the fascinating effect of number table transformations.