% --- IMPORTS ---
\documentclass[leqno]{article}
\usepackage{graphicx}
\usepackage{dsfont}
\usepackage{mathtools}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{amsthm}
\usepackage{float}
\usepackage{ragged2e}
\usepackage{tensor}
\usepackage{fancyhdr}
\usepackage{empheq}
\usepackage[most]{tcolorbox}
\usepackage{enumitem}
\usepackage{dirtytalk}
\usepackage{marginnote}
\usepackage{microtype}
\usepackage[
  a4paper,
  left=0.8in,
  right=1in,
  top=1in,
  bottom=0.5in,
  headheight=14pt,
  headsep=0.4in
]{geometry}
\setlength{\parindent}{0cm}
% --- TikZ Packages ---
\usepackage{pgfplots}
\pgfplotsset{compat=1.18}
\usepgfplotslibrary{fillbetween, polar}
\usetikzlibrary{calc, positioning}
\usetikzlibrary{angles, quotes}
\usetikzlibrary{arrows.meta}
\usetikzlibrary{decorations.pathreplacing, decorations.markings}
\usetikzlibrary{patterns}
\usetikzlibrary{intersections}
% --- URL Packages ---
\usepackage[colorlinks=true,linkcolor=red,citecolor=magenta,urlcolor=black]{hyperref}%
\urlstyle{same}
% --- END OF IMPORTS ---

% --- CUSTOM COMMANDS ---
\RenewDocumentCommand{\marks}{o m}{%
  \IfValueTF{#1}%
    {\marginnote{\raggedright\textbf{[#2]}}[#1]}%
    {\marginnote{\raggedright\textbf{[#2]}}}%
}
\newcommand{\vspacerule}[2]{%
  \par\vspace*{#1}%
  \noindent\hrulefill\par
  \vspace*{#2}%
}
\definecolor{scarlet}{RGB}{205, 40, 75}
\newcommand{\questionitem}{%
  \item\phantomsection
  \addcontentsline{toc}{section}{Question \number\value{enumi}}%
}
\renewcommand{\contentsname}{Questions}
% --- END OF CUSTOM COMMANDS ---

% --- HEADER CONFIG ---
\pagestyle{fancy}
\fancyhead{}
\fancyfoot{}
\renewcommand{\headrulewidth}{0.9pt}
\lhead{Proof by Induction - Series}
\chead{}
\rhead{\href{https://danieljsmith.org}{danieljsmith.org}}
% --- END OF HEADER CONFIG ---

\begin{document}
\vspace*{20pt}
\tableofcontents
\newpage
\begin{enumerate}[
  label=\textbf{\arabic*.},
  leftmargin=*,
  topsep=0pt,
  partopsep=0pt,
  parsep=0pt
]


% ---- Question 1 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q1]

\begin{enumerate}[label=\textbf{(\alph*)}]
  \item Prove by induction that, for all integers $n \ge 1$
  \[
  \sum_{r=1}^{n} r^3 = \frac{1}{4}n^2(n+1)^2
  \]
  \marks[-20pt]{4}
  \item Hence, or otherwise, write down a factorised expression for the sum of the first $2n$ cubes
  \[
  1^3 + 2^3 + 3^3 + \cdots + (2n)^3
  \]
  \marks[-20pt]{1}
  \item Use the formula in part (a) to write down a factorised expression for the sum of the first $n$ even cubes
  \[
  2^3 + 4^3 + 6^3 + \cdots + (2n)^3
  \]
  \marks[-20pt]{2}
  \item Hence, or otherwise, show that
  \[
  1^3 + 3^3 + 5^3 + \cdots + (2n-1)^3 = an^2(bn^2-1)
  \]
  where $a$ and $b$ are rational numbers to be determined. \marks{3} \\
\end{enumerate}

\newpage

% ---- Question 2 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q2]

Prove by induction that, for $n \geq 1$
\[
\sum_{r=1}^{n} r\,2^r = 2 + (n-1)2^{n+1}
\]
\marks[-20pt]{5}


\newpage


% ---- Question 3 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q3]

Prove by induction that for all positive integers $n$
\[
\sum_{r=1}^{n} \log\left(1-\frac{1}{4r^2}\right)=\log\left(\frac{(2n)!(2n+1)!}{2^{4n}(n!)^4}\right)
\]
\marks[-30pt]{6}
\newpage
% ---- Question 4 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q4]

Prove by induction that, for $n \in \mathbb{Z}^{+}$,
\[
\sum_{r=1}^{n} \frac{1}{r(r+1)(r+2)} = \frac{n(n+3)}{4(n+1)(n+2)}
\]
\marks[-30pt]{5}

\newpage

% ---- Question 5 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q5]

Prove by induction that, for $n \geq 1$
\[
\sum_{r=1}^{n} \left(2^r + 2 \times 3^r\right) = 2^{n+1} + 3^{n+1} - 5
\]
\marks[-20pt]{5}

\newpage

% ---- Question 6 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q6]

Prove by induction that for $n \in \mathbb{Z}^{+}$
\[
\sum_{r=1}^{n} r3^{r-1} = \frac{1 + (2n - 1)3^n}{4}
\]
\marks[-20pt]{6}

\newpage

% ---- Question 7 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q7]

\begin{enumerate}[label=\textbf{(\alph*)}]
  \item Prove by induction that, for all positive integers $n$
  \[
  \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)}=\frac{n}{2n+1}
  \]
  \marks[-30pt]{6}
  \item Hence, show that, for all positive integers $n$
  \[
  \sum_{r=n+1}^{2n} \frac{1}{(2r-1)(2r+1)}=\frac{n}{(an+b)(cn+d)}
  \]
  where $a$, $b$, $c$ and $d$ are integers to be determined. \marks{3} \\
\end{enumerate}

\newpage

% ---- Question 8 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q8]

Prove by induction that, for $n \ge 1$
\[
\sum_{r=1}^{n} \frac{r}{2^r} = 2 - \frac{n+2}{2^n}
\]
\marks[-20pt]{5}

\newpage

% ---- Question 9 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q9]

\begin{enumerate}[label=\textbf{(\alph*)}]
  \item Prove by induction that for all positive integers $n$
  \[
  \sum_{r=1}^{n} (2r-1)^3 = n^2(2n^2-1)
  \]
  \marks[-20pt]{6}
  \item Given that
  \[
  \sum_{r=1}^{2n} (2r-1)^3 = k\sum_{r=1}^{n} (2r-1)^3
  \]
  show that
  \[
  n^2 = \frac{k-4}{2(k-16)}
  \]
  \marks[-20pt]{5}
\end{enumerate}

\newpage

% ---- Question 10 ----
\questionitem
% [Source: _QBT__Proof_by_Induction___Series, Q10]

Prove by induction that for all positive integers $n$,
\[
\sum_{r=1}^{n} r(r+1)(r+2)=\frac{1}{4}n(n+1)(n+2)(n+3)
\]
\marks[-20pt]{6}


\end{enumerate}
\end{document}
