TeX source:

\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{k}^{4}=\frac{1}{5} \mathrm{n}^{5}+\frac{1}{2} \mathrm{n}^{4}+\frac{4}{12} \mathrm{n}^{3}+0 \cdot \mathrm{n}^{2}-\frac{1}{30} \mathrm{n}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)\left(3 \mathrm{n}^{2}+3 \mathrm{n}-1\right)}{30}