\(\displaystyle \sum_{r = 1}^{n} r = \dfrac 12n \:.\: (n + 1)\)

\(\displaystyle \sum_{r = 1}^{n} r^2 = \dfrac 16n \:.\: (n + 1) \:.\: (2n + 1)\)

\(\displaystyle \sum_{r = 1}^{n} r^3 = \dfrac 14n^2 \:.\: (n + 1)^2\)