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\title{Homework \#4}
\author{Assigned: February 12, 2020; Due: March 4, 2020
}
\begin{document}
\maketitle

\begin{enumerate}
\item Recall on February 6 in class we discussed
$$e^{0} + e^{2 \pi i/n} + e^{4 \pi i/n} + \cdots + e^{2(n-1) \pi i/n} = 0$$
and in order to explain why it was true we needed to show that the sum of the
real parts equals $0$ and the sum of the imaginary parts is equal to $0$.
\begin{enumerate}
\item In class I showed the following identity for $n$ even using the fact that
$\sin(2 \pi - x) = - \sin(x)$:

$$\sin(0) + \sin(2\pi/n) + \sin(4\pi/n) + \cdots + \sin(2(n-1)\pi/n) = 0$$

Do the same thing for $n$ odd (make sure it is clear, at least to yourself,
why the argument is slightly different for $n$ even and $n$ odd).
\item
Using the identity $\cos(x) = - \cos(x + \pi)$, show that
$$\cos(0) + \cos(2\pi/n) + \cos(4\pi/n) + \cdots + \cos(2(n-1)\pi/n) = 0$$
for $n$ even.
\item  Why does the same proof not work for $n$ odd ?
Show and explain what goes wrong for the example of $n=3$.
\end{enumerate}
\item Prove the following identity for $n \geq 1$ by induction on $n$.
$$
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = \frac{2^{n+1}-n-2}{2^n}~.
$$
Prove the same identity using the method of telescoping sums.
\item Let $g_1(n) := \frac{1}{1 \cdot2} + \frac{1}{2 \cdot 3} + \frac{1}{3\cdot 4} +\cdots
+ \frac{1}{n(n+1)}$.
\begin{enumerate}
\item Conjecture and prove a formula for $g_1(n)$ for $n\geq 1$.
\item Now let $g_2(n) := \frac{1}{1 \cdot3} + \frac{1}{3 \cdot 5} + \frac{1}{5\cdot 7} +\cdots
+ \frac{1}{(2n-1)(2n+1)}$. Conjecture and prove a formula for $g_2(n)$ for $n\geq 1$.
\item Define an expression $f_k(n)$ for any $k \geq 1$ so that the formula
for $f_k(n)$ agrees $f_1(n) = g_1(n)$ and $f_2(n) = g_2(n)$.  Conjecture an expression
for the value of this sum and prove it by induction.
\end{enumerate}
\end{enumerate}

\end{document}