Example 2. Find the sum $S_{n}=1+(1+2)+(1+2+3)$ $+\cdots+(1+2+3+\cdots+n)$.
Solve by forming a triangular array and then completing it to a square array:
$$
\begin{array}{c}
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
1+2+3+4+\cdots+n \\
\cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\
1+2+3+4+\cdots+n
\end{array}
$$

Except for the elements on the diagonal, the sum of the numbers in the upper right triangle is twice the sum of the numbers in the lower left triangle:
$$
2=2 \cdot 1, \quad 3+3=2(1+2), \quad 4+4+4=2(1+2+3), \cdots. 
$$
Let the sum of the numbers in the lower triangle be $x$, then
$$
\begin{aligned}
n & (1+2+\cdots+n)=x+(1+2+\cdots+n) \\
& +2 x, \\
x & =\frac{1}{3}(n-1)(1+2+\cdots+n) . \\
\therefore S_{n} & =x+(1+2+\cdots+n) \\
& =\frac{1}{3}(n-1) \frac{n(n+1)}{2}+\frac{n(n+1)}{2} \\
& =\frac{n(n+1)}{6} .
\end{aligned}
$$
(Created by: Gongjinshi Baodi County Research Office, Tianjin City)