$ \sum_{n=1}^N 1 = \dfrac11N $
$ \sum_{n=1}^N n = \dfrac12N\left(N+1\right) $
$ \sum_{n=1}^N n\left(n+1\right) = \dfrac13N\left(N+1\right)\left(N+2\right) $
$ \quad\vdots $
