TeX source:

\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{3}=\frac{1}{4} \mathrm{n}^{4}+\frac{1}{2} \mathrm{n}^{3}+\frac{3}{12} \mathrm{n}^{2}+0 \cdot \mathrm{n}=\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^{2}