Proof from Discrete Math (WIP)

gistfile1.tex

\documentclass{article}
\usepackage{amsmath}% http://ctan.org/pkg/amsmath

\everymath{\displaystyle}
\begin{document}
\Huge
$\sum_{j=0}^{n}{j\left(
    \begin{array}{c}
    n \\ 
    j
    \end{array}
\right)} = n2^{n-1}$
\newline
\newline

Now we prove it for n+1.

$\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n + 1 \\
        j
    \end{array}
\right)} = (n + 1)2^{n}$


$\sum_{j=0}^{n+1}{j\left(
	\left(
		\begin{array}{c}
    		n \\
            j
    	\end{array}
    \right)
        +
    \left(
    	\begin{array}{c}
    		n \\
        	j - 1
    	\end{array}
    \right)
\right)}$

$\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j
    \end{array}
\right)}
+
\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}$

$\sum_{j=0}^{n}{j\left(
	\begin{array}{c}
    	n \\
        j
    \end{array}
\right)}
+
\left(n+1\right)
\left(
	\begin{array}{c}
    	n \\
        n + 1
    \end{array}
\right)
+
\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}$



% Expand the combination
$\sum_{j=0}^{n}{j\left(
	\begin{array}{c}
    	n \\
        j
    \end{array}
\right)}
+
\left(
	\frac{\left(n + 1\right)n!}{\left(n+1\right)!\left(n - n + 1\right)!}
\right)
+
\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}$

% Cancel the terms
$\sum_{j=0}^{n}{j\left(
    \begin{array}{c}
    	n \\
        j
    \end{array}
\right)}
+
\left(
	\frac{n!}{n!}
\right)
+
\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}$

% Replace the terms
$n2^{n-1}
+
1
+
\sum_{j=0}^{n+1}{j\left(
	\begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}$

$n2^{n-1}
+
1
+
\sum_{j=0}^{n}{j\left(
    \begin{array}{c}
    	n \\
        j - 1
    \end{array}
\right)}
+
\left(n + 1\right)
\left(
    \begin{array}{c}
        n \\
        n + 1 - 1
    \end{array}
\right)$

% Replace some more terms
$n + 2 + n2^{n-1}
+
\sum_{j=0}^{n}{j\left(
    \begin{array}{c}
        n \\
        j - 1
    \end{array}
\right)}$

% Group terms
$2 + n\left(1 + 2^{n-1}\right)
+
\sum_{j=0}^{n}{j\left(
    \begin{array}{c}
        n \\
        j - 1
    \end{array}
\right)}$
\end{document}