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\begin
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\documentclass[a4paper,12pt]{article}
\usepackage{tma}
\myname{Peter McFarlane}
\mypin{A1234567}
\mycourse{L101}
\mytma{01}
%\marginnotes
\begin{document}
\rule[0.4pt-1em]{0.4pt}{1em}\hrulefill\rule[0.4pt-1em]{0.4pt}{1em}
\begin{center}
\begin{minipage}{0.9\linewidth}
This document is provided as an example of how to use the \textbf{tma} package.
With the 4\nth October 2013 version of the package there have been two options added to allow selection of
different numbering styles since the module M381 differs from the norm.
To use the options start your document with:
\begin{verbatim}
\documentclass[a4paper,12pt]{article}
\usepackage[OPTION]{tma}
\myname{...
\end{verbatim}
Where \textbf{OPTION} is one of the following
\begin{description}
\item[{[}roman{]}] Questions numbered as 1, 1(i), 1(i)(a)\dots
\item[{[}alph{]}] \emph{Default} Questions numbered as 1, 1(a), 1(a)(i)\dots
\end{description}
\end{minipage}
\end{center}
\rule{0.4pt}{1em}\hrulefill\rule{0.4pt}{1em}
\begin{question}
\qpart
We have $1=10^0$ and $1+2+3+4=10^1$. Prove that there are no other powers of ten which are the sum of the first $n$ integers.
We have:
\begin{gather*}
\sum_{i=1}^n i=\frac{(n)(n+1)}{2}
\intertext{Let}
\frac{(n)(n+1)}{2}=10^x\\
\Rightarrow (n)(n+1)=2^{x+1}5^x
\end{gather*}
Now, either $n$ is odd, or $n+1$ is odd.
Consider the case where $n$ is odd:\\
By the Fundamental Theorem of Arithmetic, $n$ can only have the prime factors 2 or 5. Since it is odd, it can only be a perfect power of 5. Now, $n+1$ also can only have the prime factors of 2 or 5. If $n$ is divisible by 5, then $n+1$ is not divisible by 5. Therefore $n+1$ is a perfect power of 2. Therefore:
\begin{gather*}
n=5^x\qquad \text{and}\qquad n+1=2^{x+1}\\
\Rightarrow x=0
\intertext{(for any higher $x$, $5^x\gg 2^{x+1}$)}\\
\Rightarrow n=1
\end{gather*}
Now consider the case where $n+1$ is odd:\\
By similar arguments to above, $n+1$ must be a perfect power of 5 and $n$ must be a perfect power of 2.
\begin{gather*}
n=2^{x+1}\qquad \text{and}\qquad n+1=5^{x}\\
\Rightarrow x=1
\intertext{(for any higher $x$, $5^x\gg 2^{x+1}$)}\\
\Rightarrow n=4
\end{gather*}
Therefore $n=1$ and $n=4$ are the only solutions to the original problem. \hfill $\square$\\[1cm]
\qpart[3]
\qsubpart
Show that:
\begin{gather*}
\sum_{x=1}^{n}x(x+1)=\frac{n(n+1)(n+2)}{3}\\
\end{gather*}
Let
\begin{gather*}
f(n)=\frac{n(n+1)(n+2)}{3}\\
\end{gather*}
Now, adding the n+1 term to the above
\begin{gather}\notag
\begin{aligned}
f(n)+(n+1)(n+2)&=\frac{n(n+1)(n+2)}{3}+(n+1)(n+2)\\
&=\frac{\left(n^3+3n^2+2n+\right)}{3}+n^2+3n+2\\
&=\frac{\left(n^3+6n^2+11n+6\right)}{3}\\
&=\frac{\left((n+1)(n+2)(n+3)\right)}{3}\\
&=f(n+1)
\end{aligned}
\end{gather}
Therefore, if $f(n)$ is valid, then so is $f(n+1)$.
Since $1\times2=2=\frac{1\times2\times3}{3}=f(1)$, then $f(n)$ is valid for all $n\geq1$. \hfill $\square$\\[1cm]
\qsubpart
Show that:
\begin{gather*}
\sum_{x=1}^{n}x^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\\
\end{gather*}
Let
\begin{gather*}
f(n)=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}\\
\end{gather*}
Now, adding the n+1 term to the above
\begin{align}
f(n)+(n+1)^4&=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}+(n+1)^4\notag \\
&=\frac{1}{30}(6n^5+15n^4+10n^3-n)+(n^4+4n^3+6n^2+4n+1)\notag \\
&=\frac{1}{30}(6n^5+15n^4+10n^3-n+30n^4+120n^3+180n^2+120n+30)\notag \\
&=\frac{1}{30}(6n^5+45n^4+130n^3+180n^2+119n+30)\label{a}
\end{align}
Now,
\begin{align}
f(n+1)&=\frac{(n+1)\big((n+1)+1\big)\big(2(n+1)+1\big)\big(3(n+1)^2+3(n+1)-1\big)}{30}\notag \\
&=\frac{1}{30}(n+1)(n+2)(2n+3)(3n^2+9n+5)\notag \\
&=\frac{1}{30}(6n^5+45n^4+130n^3+180n^2+119n+30)\label{b}
\end{align}
Comparing equation~\eqref{a} with equation~\eqref{b} we see that
\begin{gather}\notag
f(n)+(n+1)^4=f(n+1)
\end{gather}
Therefore, if $f(n)$ is valid, then so is $f(n+1)$.
Since $1^4=1=\frac{1\times2\times3\times5}{30}=f(1)$, then $f(n)$ is valid for all $n\geq1$.\hfill $\square$
\end{question}
\begin{question}[3]
Find the general solution of the equation
\begin{equation}
\label{e}
3\deriv{^2y}{x^2}+4\deriv{y}{x}+y=x^2
\end{equation}
The auxillary equation is
\begin{equation}
3\lambda^2+4\lambda+1=0
\end{equation}
which factorises to
\begin{equation}
(\lambda+1)(3\lambda+1)=0
\end{equation}
and so has solutions
\begin{equation}
\lambda=-1 \mbox{\ and\ }\lambda=-\frac{1}{3}
\end{equation}
As both roots are real and distinct, the complementary function is
\begin{equation}
\label{d}
y_c=C\e^{-x}+D\e^{-\frac{1}{3}x}
\end{equation}
Now, let us find the particular integral. As the right hand side of equation \ref{e} is $x^2$, our trial solution is the polynomial
\begin{equation}
y_p=px^2+qx+r
\end{equation}
\begin{equation}
\Rightarrow \deriv{y_p}{x}=2px+q
\end{equation}
\begin{equation}
\Rightarrow \deriv{^2y_p}{x^2}=2p
\end{equation}
Substituting the trial particular integral into equation \ref{e}
\begin{equation}
6p+8px+4q+px^2+qx+r=x^2
\end{equation}
\begin{equation}
\Rightarrow px^2+(8p+q)x+(6p+4q+r)=x^2
\end{equation}
\begin{equation}
\Rightarrow p=1,\ \ q=-8,\ \ r=26
\end{equation}
Thus the particular integral is
\begin{equation}
\label{c}
y_p=x^2-8x+26
\end{equation}
and combining equation \ref{d} with equation \ref{c}, by the rule of superposition, we get the general solution of equation \ref{e} to be
\begin{equation}
y=C\e^{-x}+D\e^{-\frac{1}{3}x}+x^2-8x+26
\end{equation}
\end{question}
\end{document}