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\newtheorem*{thm}{Theorem}
\theoremstyle{definition}
\begin{document}

\begin{thm}
For any $n\in \mathbb{N}$,
$$1+2+\cdots + n = \frac{n(n+1)}{2}$$
\end{thm}

\begin{proof}[Proof 1:]
We will prove the theorem by induction.  

Base case: Since $1 = 1(1+1)/2$, the theorem holds for $n=1$.

Inductive step: Assume $\sum_{k=1}^{n-1}k = \frac{(n-1)n}{2}$.  Then
\begin{eqnarray*}
1+2+ \cdots +(n-1) + n &=& \frac{(n-1)n}{2} + n \\
&=&\frac{n(n-1)}{2}+\frac{2 n}{2}\\
&=&\frac{n(n+1)}{2}.
\end{eqnarray*}

By induction, it follows that for every natural number $n$,  

$$\sum_{k=1}^{n}k = \frac{n(n+1)}{2}.$$

\end{proof}


\begin{proof}[Proof 2:]
We will prove this by reorganizing the terms of twice the sum on the left hand side.  We can reorganize the sum

\begin{align*}
2(1+2+\cdots+n) 	&=& (&1& &+& &2& &+& &\cdots& &+& &n)& +   \\
				&& (&n& &+& &n-1& &+& &\cdots& &+& &1)&  \\
				&=& (&n+1& &+& &n+1& &+& &\cdots& &+& &n+1)& \\
				&=& &n(n+1).&
\end{align*}

Dividing by 2 gives the desired equality

$$1+2+\cdots + n = \frac{n(n+1)}{2}.$$

\end{proof}

\end{document}