[tex]\displaystyle S = 1+3+6+10+...+\dfrac{n(n+1)}{2}\\ \\ S = \sum\limits_{k=1}^{n}\Big(\sum\limits_{i=1}^{k}i\Big) = \sum\limits_{k=1}^n \dfrac{k(k+1)}{2} = \dfrac{1}{2}\sum\limits_{k=1}^{n}k(k+1)\\ \\\\\\ k^3-(k+1)^3 = k^3-(k^3+3k^2+3k+1) = -(3k^2+3k+1) = \\ =-3k(k+1) - 1\Bigg|\dfrac{1}{2}\sum\limits_{k=1}^n\Big(\,\Big) \\ \\\\ -3S-\dfrac{n}{2} = \dfrac{1}{2}\sum\limits_{k=1}^n\Big[k^3-(k+1)^3\Big] \\ \\ 6S+n = \sum\limits_{k=1}^n \Big[(k+1)^3-k^3\Big] \\ \\ 6S+n = 2^3+3^3+...+(n+1)^3-1^3-2^3-...-n^3 \\ \\ 6S+n = (n+1)^3-1^3 \\ \\ 6S = (n+1)^3-(n+1)[/tex]
[tex]S = \dfrac{(n+1)^3-(n+1)}{6} = \dfrac{(n+1)\Big[(n+1)^2-1\Big]}{6} \\ \\ \Rightarrow \boxed{S = \dfrac{n(n+1)(n+2)}{6}}[/tex]
