A simple LaTex template for Math homeworks.

mathy.tex

% Resources: 
% http://www-math.mit.edu/~psh/exam/examdoc.pdf
% https://www.math.uni-bielefeld.de/~rost/amslatex/doc/amsthdoc.pdf

\documentclass[addpoints,answers,12pt]{exam} % exam class with 12 point type
\usepackage[T1]{fontenc}                     % replace default font encoding (OT1)
\usepackage{tgschola}                        % font used in the Book of Proof
\usepackage{amsmath,amsthm,amssymb,amsfonts} % packages for mathematical typesetting
\usepackage{pdfpages}                        % include pdf pages with \includepdf{dir/of/page.pdf}
\usepackage[makeroom]{cancel}                % display expressions as cancelled 


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% START CONFIGURATION
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% header and footer style
\pagestyle{headandfoot}
\firstpageheadrule      % put horizontal line on header of 1st page
\runningheadrule        % put horizontal line on header after 1st page
\firstpageheader{\student}{\subject \mbox{ ---} \assignment}{\dt}
\runningheader{\student}{\subject \mbox{ ---} \assignment}{\dt}
\firstpagefooter{}{}{} % put nothing on footer of 1st page
\runningfooter{}{}{}   % put nothing on footer after 1st page

% set some set symbols
\newcommand{\N}{\mathbb N}
\newcommand{\Z}{\mathbb Z}
\newcommand{\Q}{\mathbb Q}
\newcommand{\R}{\mathbb R}
\newcommand{\C}{\mathbb C}

% redefine tombstone symbol used at the of proofs
\renewcommand{\qedsymbol}{$\blacksquare$}

% (uncomment to) place each section on its own page
%\let\stdsection\section\renewcommand\section{\newpage\stdsection}

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% START CONFIGURATION
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% ASSIGNMENT INFO 
% (Update accordingly!)
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\newcommand{\student}{John Doe}
\newcommand{\subject}{Math Class}
\newcommand{\assignment}{Homework 00}
\newcommand{\dt}{\today}

\begin{document}
\begin{questions}

\question{Prove that $ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $ for $ n \ge 0 $, by induction.}
\begin{solution}
  \begin{proof}\hfil
  
  \textbf{Base case.} When $  n = 0 $, LHS = $ 0 $ and RHS = $ \frac{0\cdot 1}{2} = 0$. Thus RHS = LHS.
  \textbf{Inductive step.} We assumed that $ \sum_{i=1}^{q} i = \frac{q(q+1)}{2} $ is true for an arbitrary fixed integer $ q $ and attempt to prove the validity of the formula for $ (q+1) $. Thus 
  \begin{align*}
  \sum_{i=1}^{q+1} i &=  \underbrace{\sum_{i=1}^{q} i}_{\frac{q(q+1)}{2}} + (q+1) \\
  &= \frac{q(q+1)}{2} + (q+1) \\
  &= \frac{q(q+1) + 2(q+1)}{2} \\
  &= \frac{(q+1)(q+1)}{2} \mbox{.}
  \end{align*} 
  which is precisely the right-side of $ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $ when $ n = (q+1) $.
  Therefore, by the principle of mathematical induction, $ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} $.
  \end{proof}
\end{solution}

\end{questions}
\end{document}