Question

If $\sum _{r=1}^{n}r=55,\mathrm{find}\sum _{r=1}^n{r}^3$.

Answer 1

$$\begin{gathered} \sum _{r=1}^n{r}^3=1^3+2^3+3^3+...+n^3 \\ ={\left[\frac{n\left(n+1\right)}{2}\right]}^2 \\ ={\left[\sum _{r=1}^{n}r\right]}^2 \\ ={\left[55\right]}^2 \\ =3025 \end{gathered}$$