\begin{frame} \small \begin{example} We use induction to prove that: \vspace{-1ex} \begin{align*} 1 + 2 + \ldots + n = \frac{n \cdot (n+1)}{2} \end{align*} \vspace{-5ex} \begin{enumerate} \item<2-> Base case $n = 0$: \onslide<3-> Then \vspace{-1ex} \begin{align*} 1 + 2 + \ldots + 0 &= 0 & \text{ and }&& \frac{0 \cdot (0+1)}{2} &= 0 \end{align*} \vspace{-2ex} \onslide<4-> Thus the statement holds for $n = 0$. \medskip \item<5-> Induction step:\\ \onslide<6-> \alert<6->{Induction hypothesis (IH): Assume the statement hold for $n$.}\\[1ex] \onslide<7-> We show it for $n+1$: \vspace{-1ex} \begin{align*} 1 + 2 + \ldots + n& + (n+1) \onslide<8->{= \frac{n \cdot (n+1)}{2} + (n+1) \quad \text{by \alert{IH}}}\\ &\onslide<9->{= \frac{n \cdot (n+1) + 2\cdot(n+1)}{2}} \onslide<10->{= \frac{(n+1) \cdot ((n+1) + 1)}{2}} \end{align*} \vspace{-5ex} \end{enumerate} \onslide<11-> Hence the formula holds for all $n \in \NN$. \end{example} \end{frame}