\[ \begin{split}
&( \frac{21! \cdot 2^3}{3! 18! \cdot 4^{18}} +
\frac{21! \cdot 2^2}{2! 2! 17! \cdot 5^2 \cdot 4^{17}} \\
& + \frac{21! \cdot 2}{4! 16! \cdot 5^4 \cdot 4^{16}} +
frac{21!}{6! 15! \cdot 5^6 \cdot 4^{15}} ) x^{36} \\
&= ( \frac{38138681}{67108864000000} ) x^{36}
\end{split} \]
Is there anything wrong with it? The log doesn't tell me anything useful. Here is the full file:
\documentclass[letterpaper, 12pt, titlepage]{article}
\pagestyle{myheadings} \markright{Fakaff \hfill MATH2P81 \hfill}
\usepackage{amsmath}
\begin{document}
\title{MATH 2P81 \\ Assignment \#1}
\date{\today}
\author{Fakaff \\ \texttt{4847653}}
\maketitle
\begin{enumerate}
\item All terms of the expansion will be of the form $ \binom{21}{a, b, c} 2^a x^b (x^2)^c $. To
find the coefficient of $ x^{36} $, we must first find all sets $ \{a, b, c\} $ such that
$ a + b + c = 21 $ and $ a + b + 2c = 36 $:
$\{3, 0, 18\} $
$ \{2, 2, 17\} $
$ \{1, 4, 16 \} $
$ \{0, 6, 15 \} $
Plugging these numbers into our formula, we get:
\[ \begin{split}
&( \frac{21! \cdot 2^3}{3! 18! \cdot 4^{18}} + \frac{21! \cdot 2^2}{2! 2! 17! \cdot 5^2 \cdot 4^{17}} \\
& + \frac{21! \cdot 2}{4! 16! \cdot 5^4 \cdot 4^{16}} + frac{21!}{6! 15! \cdot 5^6 \cdot 4^{15}} ) x^{36} \\
&= ( \frac{38138681}{67108864000000} ) x^{36}
\end{split} \]
Checking in maple by using the \texttt{expand( (2 - (x/5) + (x^2/4 )^21 )} command confirms this result.
\end{enumerate}
\end{document}
