logarytm $\log_28=3 $
suma \[ \sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6} \]
granica $ \lim_{x\to 0} $
Alfabet grecki $ \alpha,\beta, \omega, \Gamma, \Delta, \Omega $
Liczba e \[ e= \lim_{n\to 0} \left( 1+\frac{1}{n} \right)^n \]
$$\sum_{i=0}^n i^2 = \frac{(n^2+n)(2n+1)}{6}$$
\begin{equation}
E = mc^2
\end{equation}
\begin{equation*} e^{\pi i} + 1 = 0 \end{equation*}