$\displaystyle \sum_{k=1}^n k = \dfrac{1}{2}n(n+1)$

$\displaystyle \sum_{k=1}^n k^2 = \dfrac{1}{6}n(n+1)(2n+1)$

$\displaystyle \sum_{k=1}^n k^3 = \left\{ \dfrac{1}{2}n(n+1) \right\}^2$