Well, known triangular numbers $$1, 3, 6, 10, 15, 21, 28, 36, 45, \cdots$$ I am looking for different methods to get the sum of those numbers. I know we can write it as below
$$\displaystyle\sum\limits\_{i=1}^n \displaystyle\sum\limits\_{k=1}^i i\\=
\displaystyle\sum\limits\_{i=1}^n \frac{i(i+1)}{2}=\\
\frac 12 \displaystyle\sum\limits\_{i=1}^n (i+i^2)=\\
\frac12 (\frac{n(n+1)}{2}+\frac{n(n+1)(2n+1)}{6})=\\
\frac{n(n+1)(n+2)}3$$ I know we that we can show it by $\mathit{mathematical} \space \mathscr{Induction}$, But are there more methods to show this fact? As an example a proof without a word, or something else. Thanks in advance for any hint, solution, or different ideas. Also, I am looking for specifics about the heuristic solution.