User talk:Crisófilax

Se puede ver asi $\begin{matrix}\sum_{k=1}^n k^2\end{matrix}$ o asi $\sum_{k=1}^n k^2$ .

$\sum_{1\le k\le n \atop (k,n)=1}\phi(k) = \frac{1}{2}n\phi(n)$ for $\;n>1$

$\sum_{r \in \mathbb{Q}^+}\frac{r^2}{2^{\text{num}(r)}-1}=\pi^2$