Math 302 HW 0 - B. Rodriguez

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\begin{document}
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\noindent
\textbf{MATH 302-01} \hfill \textbf{Breanna Rodriguez} \\
\normalsize Prof. Soto \hfill Due Date: 08/25/19 \\

% This is where you name your homework
\begin{center}
\textbf{Homework 0}
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% This is how you call the environment for the statement to be proved.
\begin{statement}{\#1}
Prove that    
\begin{equation}
\label{bree}
    1 + 2 + 3+...+ n = \sum_{k=1}^n k = \frac{n(n+1)}{2}.
\end{equation}
\end{statement}

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\begin{proof}
We would like to prove that $ \sum_{k=1}^n k = \frac{n(n+1}{2}$ so we proceed by induction. \\

\textbf{Base Case}. For $n = 1$ we see that the left hand side of (\ref{bree}) is 1 whereas the right hand side is given by 
$$ \frac{1(1+1)}{2}= 1. $$
as well. Hence the statement is true for $n =1$ \\

\textbf{Induction Case}. Assume that the statement holds for n+1, that is we need to show that
\begin{align*}
 1 + 2 + 3 + \dotsc + n + (n+1) &= \frac{(n+1)((n+1)+1)}{2} \\
 & = \frac{(n+1)(n+2)}{2}
\end{align*}
Thus we will begin with the left hand side of (\ref{bree}) to reach our conclusion. By our assumption we know that 
$$ 1 + 2 + 3 + \dotsc + n + (n+1) = \frac{n(n+1)}{2} + (n+1)
$$
Thus we use a bit of algebra as follows to reach our conclusion: 
\begin{align*}
    1 + 2 + 3 + \dotsc + n+ (n+1) & = \frac{n(n+1)}{2}+ (n+1) \\ 
    & = \frac{n(n+1)}{2} + \frac{2(n+1)}{2} \\
    & = \frac{(n+1)(n+2)}{2}.
\end{align*}
Thus by induction we see that statement (\ref{bree}) is true.

\end{proof}
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