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\begin{document}

\begin{center}
{\Large Warm-Up Problems and Lecture Problems

Solutions

March 31, 2003}
\end{center}

\begin{enumerate}

\item  Find the pattern for the following sequences.  Write the formula for the $n$-th term of the sequence.  (Hint: the idea for all of them are similar.)
\begin{enumerate}
\item  $1,2,4,8, \ldots$
	\\\sol $a_n = 2^{n-1}$

\item  $\frac{1}{2}, \frac{3}{4}, \frac{9}{8}, \frac{27}{16}, \ldots$
	\\\sol $a_n = \frac{3^{n-1}}{2^n} = \left(\frac{1}{2}\right)\left(\frac{3}{2}\right)^{n-1}$

\item  $100, 20, 4, \ldots$
	\\\sol $a_n=100\left(\frac{1}{5}\right)^{n-1}$

\item  $100, -20, 4, \ldots$
	\\\sol $a_n=100\left(-\frac{1}{5}\right)^{n-1} = 100(-1)^{n-1}\left(\frac{1}{5}\right)^{n-1}$
\end{enumerate}

\item  Can you find the sum of the sequences above (using your calculator, can you estimate the sum?)
	\\\sol These are easiest thought of in what we will study today.
	These are all geometric series.
	Here are the sums:
	\begin{enumerate}
	\item  \sol
		\[ 
		1 + 2 + 4 + 8 + \cdots = \sum_{n=1}^\infty 2^{n-1} = \infty
		\]

	\item  \sol
	\[
	\frac{1}{2}, + \frac{3}{4} + \frac{9}{8} + \frac{27}{16} + \cdots
	= \sum_{n=1}^\infty \left(\frac{1}{2}\right)\left(\frac{3}{2}\right)^{n-1}
	= \infty
	\]

	\item  \sol
	\[
	100 + 20 + 4 + \cdots
	= 
	\sum_{n=1}^\infty 100\left(\frac{1}{5}\right)^{n-1}
	= 125
	\]

	\item  \sol
	\[
	100 - 20 + 4 + \cdots
	= \sum_{n=1}^\infty 100(-1)^{n-1}\left(\frac{1}{5}\right)^{n-1}
	= \frac{250}{3}
	\]
	\end{enumerate}

\newpage

\item  Consider the series:
	\[
	\sum_{n=1}^{\infty} \frac{1}{n+4} - \frac{1}{n+5}
	\]
	Write down the series term by term and find the sum.
	\\\sol
	\begin{eqnarray*}
	\sum_{n=1}^{\infty} \frac{1}{n+4} - \frac{1}{n+5}
	& = & 
	\left(\frac{1}{5} - \frac{1}{6}\right)
	+
	\left(\frac{1}{6} - \frac{1}{7}\right)	
	+
	\left(\frac{1}{7} - \frac{1}{8}\right)	
	+
	\left(\frac{1}{8} - \frac{1}{9}\right)	
	+
	\left(\frac{1}{9} - \frac{1}{10}\right) 
	+ \cdots \\
	& = & \frac{1}{5}
	\end{eqnarray*}


\end{enumerate}


\end{document}