${\displaystyle \sum _{r=1}^{n}r={\displaystyle \frac{1}{2}}n.(n+1)}$

${\displaystyle \sum _{r=1}^n{r}^2={\displaystyle \frac{1}{6}}n.(n+1).(2n+1)}$

${\displaystyle \sum _{r=1}^n{r}^3={\displaystyle \frac{1}{4}}{n}^2.(n+1{)}^2}$