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Mathematics LibreTexts

9.E: Sequences and the Binomial Theorem (Exercises)

9.1: Sequences

\newpage

\subsection{Exercises}

In Exercises \ref{writeoutseqfirst} - \ref{writeoutseqlast}, write out the first four terms of the given sequence.

\begin{multicols}{2}

\begin{enumerate}

\item $a_{n} = 2^{n} - 1 \vphantom{d_{j} = (-1)^{\dfrac{j(j+1)}{2}}}$, $n \geq 0$ \label{writeoutseqfirst}

\item $d_{j} = (-1)^{\dfrac{j(j+1)}{2}}$, $j \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ 5k - 2 \right\}_{k=1}^{\infty} \vphantom{\left\{ \dfrac{n^2+1}{n+1} \right\}_{n=0}^{\infty}}$

\item $\left\{ \dfrac{n^2+1}{n+1} \right\}_{n=0}^{\infty}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ \dfrac{x^{n}}{n^{2}} \right\}_{n=1}^{\infty}$

\item $\left\{ \dfrac{\ln(n)}{n} \right\}_{n=1}^{\infty} \vphantom{\left\{ \dfrac{x^{n}}{n^{2}} \right\}_{n=1}^{\infty}}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a_{\mbox{\tiny$1$}} = 3$, $a_{n+1} = a_{n} - 1$, $n \geq 1 \vphantom{d_{m} = \dfrac{d_{m\mbox{-\tiny$1$}}}{100}}$

\item $d_{\mbox{\tiny$0$}} = 12$, $d_{m} = \dfrac{d_{m\mbox{-\tiny$1$}}}{100}$, $m \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $b_{\mbox{\tiny$1$}} = 2$, $b_{k\mbox{+\tiny$1$}} =3b_{k}+1 \vphantom{\dfrac{c_{j\mbox{-\tiny$1$}}}{(j+1)(j+2)}}$, $k \geq 1$

\item $c_{\mbox{\tiny$0$}} = -2$, $c_{j} = \dfrac{c_{j\mbox{-\tiny$1$}}}{(j+1)(j+2)}$, $m \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a_{\mbox{\tiny$1$}} = 117$, $a_{n\mbox{+\tiny$1$}} = \dfrac{1}{a_{n}}$, $n \geq 1$

\item $s_{\mbox{\tiny$0$}} = 1$, $s_{n\mbox{+\tiny$1$}} = x^{n + 1} + s_{n}$, $n \geq 0$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $F_{\mbox{\tiny$0$}} = 1$, $F_{\mbox{\tiny$1$}} = 1$, $F_{n} = F_{n\mbox{-\tiny$1$}} + F_{n\mbox{-\tiny$2$}}$, $n \geq 2$ (This is the famous \href{http://en.wikipedia.org/wiki/Fibonac...line{Fibonacci Sequence}} ) \label{writeoutseqlast}

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\end{enumerate}

In Exercises \ref{alggeoneithfirst} - \ref{alggeoneithlast} determine if the given sequence is arithmetic, geometric or neither. If it is arithmetic, find the common difference $d$; if it is geometric, find the common ratio $r$.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left\{ 3n-5 \right\}_{n=1}^{\infty}$ \label{alggeoneithfirst}

\item $a_{n} = n^2+3n+2$, $n \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{1}{3}$, $\dfrac{1}{6}$, $\dfrac{1}{12}$, $\dfrac{1}{24} \vphantom{\left\{ 3 \left(\dfrac{1}{5}\right)^{n-1} \right\}_{n=1}^{\infty}}$, \ldots

\item $\left\{ 3 \left(\dfrac{1}{5}\right)^{n-1} \right\}_{n=1}^{\infty}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $17$, $5$, $-7$, $-19$, \ldots

\item $2$, $22$, $222$, $2222$, \ldots

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $0.9$, $9$, $90$, $900 \vphantom{a_{n} = \dfrac{n!}{2}}$, \ldots

\item $a_{n} = \dfrac{n!}{2}$, $n \geq 0$. \label{alggeoneithlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{nthtermfirst} - \ref{nthtermlast}, find an explicit formula for the $n^{\mbox{\scriptsize th}}$ term of the given sequence. Use the formulas in Equation \ref{arithgeoformula} as needed.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $3$, $5$, $7$, $9 \vphantom{-\dfrac{1}{8}}$, \ldots \label{nthtermfirst}

\item $1$, $-\dfrac{1}{2}$, $\dfrac{1}{4}$, $-\dfrac{1}{8}$, \ldots

\item $1$, $\dfrac{2}{3}$, $\dfrac{4}{5}$, $\dfrac{8}{7}$, \ldots

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $1$, $\dfrac{2}{3}$, $\dfrac{1}{3}$, $\dfrac{4}{27} \vphantom{\dfrac{x^7}{7}}$, \ldots

\item $1$, $\dfrac{1}{4}$, $\dfrac{1}{9}$, $\dfrac{1}{16} \vphantom{-\dfrac{x^7}{7}}$, \ldots

\item $x$, $-\dfrac{x^3}{3}$, $\dfrac{x^5}{5}$, $-\dfrac{x^7}{7}$, \ldots

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $0.9, 0.99, 0.999, 0.9999, \ldots$

\item $27, 64, 125, 216, \ldots$

\item $1, 0, 1, 0, \ldots$ \label{nthtermlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

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\item Find a sequence which is both arithmetic and geometric. (Hint: Start with $a_{n} = c$ for all $n$.)

\item Show that a geometric sequence can be transformed into an arithmetic sequence by taking the natural logarithm of the terms.

\item Thomas Robert Malthus is credited with saying, ``The power of population is indefinitely greater than the power in the earth to produce subsistence for man. Population, when unchecked, increases in a geometrical ratio. Subsistence increases only in an arithmetical ratio. A slight acquaintance with numbers will show the immensity of the first power in comparison with the second.'' (See this \href{http://en.wikipedia.org/wiki/Malthus...line{webpage}} for more information.) Discuss this quote with your classmates from a sequences point of view.

\item This classic problem involving sequences shows the power of geometric sequences. Suppose that a wealthy benefactor agrees to give you one penny today and then double the amount she gives you each day for 30 days. So, for example, you get two pennies on the second day and four pennies on the third day. How many pennies do you get on the $30^{\mbox{\scriptsize th}}$ day? What is the \underline{total} dollar value of the gift you have received?

\item Research the terms `arithmetic mean' and `geometric mean.' With the help of your classmates, show that a given term of a arithmetic sequence $a_{k}$, $k \geq 2$ is the arithmetic mean of the term immediately preceding, $a_{k\mbox{\tiny$-1$}}$ it and immediately following it, $a_{k\mbox{\tiny$+1$}}$. State and prove an analogous result for geometric sequences.

\item Discuss with your classmates how the results of this section might change if we were to examine sequences of other mathematical things like complex numbers or matrices. Find an explicit formula for the $n^{\mbox{\scriptsize th}}$ term of the sequence $i, -1, -i, 1, i, \ldots$. List out the first four terms of the matrix sequences we discussed in Exercise \ref{Markovchain} in Section \ref{MatArithmetic}.

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{2}

\begin{enumerate}

\item $0, 1, 3, 7$

\item $-1, -1, 1, 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $3, 8, 13, 18$

\item $1, 1, \frac{5}{3}, \frac{5}{2}$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

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\item $x, \frac{x^{2}}{4}, \frac{x^{3}}{9}, \frac{x^{4}}{16}$

\item $0, \frac{\ln(2)}{2}, \frac{\ln(3)}{3}, \frac{\ln(4)}{4}$

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\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $3, 2, 1, 0$

\item $12, 0.12, 0.0012, 0.000012$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $2, 7, 22, 67$

\item $-2, -\frac{1}{3}, -\frac{1}{36}, -\frac{1}{720}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $117, \frac{1}{117}, 117, \frac{1}{117}$

\item $1, x + 1, x^{2} + x + 1, x^{3} + x^{2} + x + 1 $

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $1, 1, 2, 3$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item arithmetic, $d = 3$

\item neither

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item geometric, $r = \frac{1}{2}$

\item geometric, $r = \frac{1}{5}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item arithmetic, $d = -12$

\item neither

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item geometric, $r = 10$

\item neither

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a_{n} = 1 + 2n, \; n \geq 1$

\item $a_{n} = \left(-\frac{1}{2}\right)^{n - 1}, \; n \geq 1$

\item $a_{n} = \frac{2^{n - 1}}{2n - 1}, \; n \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a_{n} = \frac{n}{3^{n - 1}}, \; n \geq 1$

\item $a_{n} = \frac{1}{n^{2}}, \; n \geq 1$

\item $\frac{(-1)^{n - 1}x^{2n - 1}}{2n -1}, \; n \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $a_{n} = \frac{10^{n} - 1}{10^{n}}, \; n \geq 1$

\item $a_{n} = (n + 2)^{3}, \; n \geq 1$

\item $a_{n} = \frac{1 + (-1)^{n-1}}{2}, \; n \geq 1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\closegraphsfile

9.2: Summation Notation

\newpage

\subsection{Exercises}

In Exercises \ref{sumfirst} - \ref{sumlast}, find the value of each sum using Definition \ref{sigmanotation}.

\begin{multicols}{4}

\begin{enumerate}

\item $\displaystyle \sum_{g = 4}^{9} (5g + 3)$ \label{sumfirst}

\item $\displaystyle \sum_{k = 3}^{8} \frac{1}{k}$

\item $\displaystyle \sum_{j = 0}^{5} 2^{j}$

\item $\displaystyle \sum_{k = 0}^{2} (3k - 5)x^{k}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle \sum_{i = 1}^{4} \frac{1}{4}(i^{2} + 1)$

\item $\displaystyle \sum_{n = 1}^{100} (-1)^{n}$

\item $\displaystyle \sum_{n = 1}^{5} \frac{(n+1)!}{n!}$

\item $\displaystyle \sum_{j = 1}^{3} \frac{5!}{j! \, (5-j)!}$ \label{sumlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{writesumfirst} - \ref{writesumlast}, rewrite the sum using summation notation.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $8 + 11 + 14 + 17 + 20$ \label{writesumfirst}

\item $1 - 2 + 3 - 4 + 5 - 6 + 7 - 8$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7}$

\item $1 + 2 + 4 + \cdots + 2^{29} \vphantom{x - \dfrac{x^{3}}{3} + \dfrac{x^{5}}{5} - \dfrac{x^{7}}{7}}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $2 + \frac{3}{2} + \frac{4}{3} + \frac{5}{4} + \frac{6}{5}$

\item $-\ln(3) + \ln(4) - \ln(5) + \cdots + \ln(20)$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $1 - \frac{1}{4} + \frac{1}{9} - \frac{1}{16} + \frac{1}{25} - \frac{1}{36}$

\item $\frac{1}{2}(x - 5) + \frac{1}{4}(x - 5)^{2} + \frac{1}{6}(x - 5)^{3} + \frac{1}{8}(x - 5)^{4}$ \label{writesumlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{findsumformfirst} - \ref{findsumformulalast}, use the formulas in Equation \ref{arithgeosum} to find the sum.

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle \sum_{n = 1}^{10} 5n+3$ \label{findsumformfirst}

\item $\displaystyle \sum_{n = 1}^{20} 2n-1$

\item $\displaystyle \sum_{k = 0}^{15} 3-k$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle \sum_{n = 1}^{10} \left(\frac{1}{2}\right)^{n}$

\item $\displaystyle \sum_{n = 1}^{5} \left(\frac{3}{2}\right)^{n}$

\item $\displaystyle \sum_{k = 0}^{5} 2\left(\frac{1}{4}\right)^{k}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $1+4+7+ \ldots +295$

\item $4+2+0-2- \ldots - 146$

\item $1+3+9+ \ldots + 2187$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{256}\vphantom{\displaystyle \sum_{n = 1}^{10} -2n + \left(\frac{5}{3}\right)^{n}}$

\item $3 - \frac{3}{2} + \frac{3}{4} - \frac{3}{8}+- \dots +\frac{3}{256} \vphantom{\displaystyle \sum_{n = 1}^{10} -2n + \left(\frac{5}{3}\right)^{n}}$

\item $\displaystyle \sum_{n = 1}^{10} -2n + \left(\frac{5}{3}\right)^{n}$ \label{findsumformulalast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{dectofracfirst} - \ref{dectofraclast}, use Theorem \ref{geoseries} to express each repeating decimal as a fraction of integers.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $0.\overline{7}$ \label{dectofracfirst}

\item $0.\overline{13}$

\item $10.\overline{159}$

\item $-5.8\overline{67}$ \label{dectofraclast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\pagebreak

In Exercises \ref{annuityfirst} - \ref{annuitylast}, use Equation \ref{fvannuity} to compute the future value of the annuity with the given terms. In all cases, assume the payment is made monthly, the interest rate given is the annual rate, and interest is compounded monthly.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item payments are \$300, interest rate is 2.5\%, term is 17 years. \label{annuityfirst}

\item payments are \$50, interest rate is 1.0\%, term is 30 years.

\item payments are \$100, interest rate is 2.0\%, term is 20 years

\item payments are \$100, interest rate is 2.0\%, term is 25 years

\item payments are \$100, interest rate is 2.0\%, term is 30 years

\item payments are \$100, interest rate is 2.0\%, term is 35 years

\label{annuitylast}

\item Suppose an ordinary annuity offers an annual interest rate of $2 \%$, compounded monthly, for 30 years. What should the monthly payment be to have $\$100,\!000$ at the end of the term?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Prove the properties listed in Theorem \ref{sigmaprops}.

\item Show that the formula for the future value of an annuity due is \[A = P(1 + i)\left[\frac{(1 + i)^{nt} - 1}{i}\right]\]

\item Discuss with your classmates what goes wrong when trying to find the following sums.\footnote{When in doubt, write them out!}

\begin{enumerate}

\begin{multicols}{3}

\item $\displaystyle{ \sum_{k=1}^{\infty} 2^{k-1}}$

\item $\displaystyle{ \sum_{k=1}^{\infty} (1.0001)^{k-1}}$

\item $\displaystyle{ \sum_{k=1}^{\infty} (-1)^{k-1}}$

\end{multicols}

\end{enumerate}

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{4}

\begin{enumerate}

\item $213$

\item $\frac{341}{280}$

\item $63$

\item $-5 - 2x + x^{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\frac{17}{2}$

\item $0$

\item $20$

\item $25$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle \sum_{k = 1}^{5} (3k + 5)$

\item $\displaystyle \sum_{k = 1}^{8} (-1)^{k - 1}k$

\item $\displaystyle \sum_{k = 1}^{4} (-1)^{k - 1} \frac{x^{}}{2k - 1}$

\item $\displaystyle \sum_{k = 1}^{30} 2^{k-1}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle \sum_{k = 1}^{5} \frac{k + 1}{k}$

\item $\displaystyle \sum_{k = 3}^{20} (-1)^{k} \ln(k)$

\item $\displaystyle \sum_{k = 1}^{6} \frac{(-1)^{k - 1}}{k^{2}}$

\item $\displaystyle \sum_{k = 1}^{4} \frac{1}{2k}(x - 5)^{k}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $305$

\item $400$

\item $-72$

\item $\dfrac{1023}{1024}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{633}{32}$

\item $\dfrac{1365}{512}$

\item $14652$

\item $-5396$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $3280$

\item $\dfrac{255}{256}$

\item $\dfrac{513}{256}$

\item $\dfrac{17771050}{59049}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{7}{9}$

\item $\dfrac{13}{99}$

\item $\dfrac{3383}{333}$

\item $-\dfrac{5809}{990}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item \$76,\!163.67

\item $\$20,\!981.40$

\item $\$29,\!479.69$

\item $\$38,\!882.12$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $49,\!272.55$

\item $60,\!754.80$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item For $\$100,\!000$, the monthly payment is $\approx \$202.95$.

\end{enumerate}

\closegraphsfile

9.3: Mathematical Induction

\subsection{Exercises}

In Exercises \ref{proofindfirst} - \ref{proofindlast}, prove each assertion using the Principle of Mathematical Induction.

\begin{enumerate}

\item $\displaystyle{ \sum_{j=1}^{n} j^2 = \dfrac{n(n+1)(2n+1)}{6}}$ \label{proofindfirst}

\item $\displaystyle{ \sum_{j=1}^{n} j^3 = \dfrac{n^2(n+1)^2}{4}}$

\item $2^{n} > 500 n$ for $n > 12$

\item $3^{n} \geq n^3$ for $n \geq 4$

\item Use the Product Rule for Absolute Value to show $\left|x^{n}\right| = |x|^{n}$ for all real numbers $x$ and all natural numbers $n \geq 1$

\item Use the Product Rule for Logarithms to show $\log\left(x^{n}\right) = n \log(x)$ for all real numbers $x > 0$ and all natural numbers $n \geq 1$.

\item $\left[ \begin{array}{cc} a & 0 \\ 0 & b \\ \end{array} \right]^{n} = \left[ \begin{array}{cc} a^{n} & 0 \\ 0 & b^{n} \\ \end{array} \right]$ for $n \geq 1$. \label{proofindlast}

\item Prove Equations \ref{arithgeoformula} and \ref{arithgeosum} for the case of geometric sequences. That is:

\begin{enumerate}

\item For the sequence $a_{\mbox{\tiny $1$}} = a$, $a_{n\mbox{\tiny{$+1$}}} = r a_{n}$, $n \geq 1$, prove $a_{n} = ar^{n-1}$, $n \geq 1$.

\item $\displaystyle{\sum_{j=1}^{n} a r^{n-1} = a \left( \dfrac{1-r^n}{1-r}\right)}$, if $r \neq 1$, $\displaystyle{\sum_{j=1}^{n} a r^{n-1} = na}$, if $r=1$.

\end{enumerate}

\item Prove that the determinant of a lower triangular matrix is the product of the entries on the main diagonal. (See Exercise \ref{triangularmatrices} in Section \ref{MatArithmetic}.) Use this result to then show $\det\left(I_{n}\right) = 1$ where $I_{n}$ is the $n \times n$ identity matrix.

\item Discuss the classic `paradox' \href{http://en.wikipedia.org/wiki/All_hor...\underline{All Horses are the Same Color}} problem with your classmates.

\end{enumerate}

\newpage

\subsection{Selected Answers}

\begin{enumerate}

\item Let $P(n)$ be the sentence $\displaystyle{ \sum_{j=1}^{n} j^2 = \dfrac{n(n+1)(2n+1)}{6}}$. For the base case, $n=1$, we get

\[ \begin{array}{rcl}

\displaystyle{ \sum_{j=1}^{1} j^2} & \stackrel{?}{=} & \dfrac{(1)(1+1)(2(1)+1)}{6} \\ [15pt]

1^2 & = & 1 \, \checkmark \\ \end{array} \]

We now assume $P(k)$ is true and use it to show $P(k+1)$ is true. We have

\[ \begin{array}{rcl}

\displaystyle{ \sum_{j=1}^{k+1} j^2} & \stackrel{?}{=} & \dfrac{(k+1)((k+1)+1)(2(k+1)+1)}{6} \\ [15pt]

\displaystyle{ \sum_{j=1}^{k} j^2} + (k+1)^2 & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\ [15pt]

\underbrace{\dfrac{k(k+1)(2k+1)}{6}}_{\text{Using $P(k)$}} + (k+1)^2 & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\

&& \\

\dfrac{k(k+1)(2k+1)}{6} + \dfrac{6(k+1)^2}{6} & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\ [10pt]

\dfrac{k(k+1)(2k+1)+6(k+1)^2}{6} & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\ [10pt]

\dfrac{(k+1)(k(2k+1)+6(k+1))}{6} & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\ [10pt]

\dfrac{(k+1)\left(2k^2+7k+6\right)}{6} & \stackrel{?}{=} & \dfrac{(k+1)(k+2)(2k+3)}{6} \\ [10pt]

\dfrac{(k+1)(k+2)(2k+3)}{6} & = & \dfrac{(k+1)(k+2)(2k+3)}{6} \, \checkmark \\ [10pt]

\end{array} \]

By induction, $\displaystyle{ \sum_{j=1}^{n} j^2 = \dfrac{n(n+1)(2n+1)}{6}}$ is true for all natural numbers $n \geq 1$.

\addtocounter{enumi}{2}

\item Let $P(n)$ be the sentence $3^n > n^3$. Our base case is $n=4$ and we check $3^4 = 81$ and $4^3 = 64$ so that $3^4 > 4^3$ as required. We now assume $P(k)$ is true, that is $3^k > k^3$, and try to show $P(k+1)$ is true. We note that $3^{k+1} = 3 \cdot 3^{k} > 3k^3$ and so we are done if we can show $3k^3 > (k+1)^3$ for $k \geq 4$. We can solve the inequality $3x^3 > (x+1)^3$ using the techniques of Section \ref{AlgebraicFunctions}, and doing so gives us $x > \frac{1}{\sqrt[3]{3}-1} \approx 2.26.$ Hence, for $k \geq 4$, $3^{k+1} = 3 \cdot 3^{k} > 3k^3 > (k+1)^3$ so that $3^{k+1} > (k+1)^3$. By induction, $3^n > n^3$ is true for all natural numbers $n \geq 4$.

\addtocounter{enumi}{1}

\item Let $P(n)$ be the sentence $\log\left(x^n \right) = n \log(x)$. For the duration of this argument, we assume $x > 0$. The base case $P(1)$ amounts checking that $\log\left(x^1\right) = 1 \log(x)$ which is clearly true. Next we assume $P(k)$ is true, that is $\log\left(x^{k}\right) = k \log(x)$ and try to show $P(k+1)$ is true. Using the Product Rule for Logarithms along with the induction hypothesis, we get

\[\log\left(x^{k+1}\right) = \log\left(x^{k} \cdot x\right) = \log\left(x^{k}\right) + \log(x) = k \log(x) + \log(x) = (k+1) \log(x) \]

Hence, $\log\left(x^{k+1}\right) = (k+1) \log(x)$. By induction $\log\left(x^n \right) = n \log(x)$ is true for all $x>0$ and all natural numbers $n \geq 1$.

\addtocounter{enumi}{2}

\item Let $A$ be an $n \times n$ lower triangular matrix. We proceed to prove the $\det(A)$ is the product of the entries along the main diagonal by inducting on $n$. For $n=1$, $A = [a]$ and $\det(A) = a$, so the result is (trivially) true. Next suppose the result is true for $k \times k$ lower triangular matrices. Let $A$ be a $(k+1) \times (k+1)$ lower triangular matrix. Expanding $\det(A)$ along the first row, we have

\[ \det(A) = \displaystyle{\sum_{p=1}^{n} a_{\mbox{\tiny$1$}p} C_{\mbox{\tiny$1$}p}} \]

Since $a_{\mbox{\tiny$1$}p} = 0$ for $2 \leq p \leq k+1$, this simplifies $\det(A) = a_{\mbox{\tiny$11$}}C_{\mbox{\tiny$11$}}$. By definition, we know that $C_{\mbox{\tiny$11$}} = (-1)^{1+1} \det\left(A_{\mbox{\tiny$11$}}\right) =\det\left(A_{\mbox{\tiny$11$}}\right)$ where $A_{\mbox{\tiny$11$}}$ is $k \times k$ matrix obtained by deleting the first row and first column of $A$. Since $A$ is lower triangular, so is $A_{\mbox{\tiny$11$}}$ and, as such, the induction hypothesis applies to $A_{\mbox{\tiny$11$}}$. In other words, $\det\left(A_{\mbox{\tiny$11$}}\right)$ is the product of the entries along $A_{\mbox{\tiny$11$}}$'s main diagonal. Now, the entries on the main diagonal of $A_{\mbox{\tiny$11$}}$ are the entries $a_{\mbox{\tiny$22$}}$, $a_{\mbox{\tiny$33$}}$, \ldots, $a_{(k\mbox{\tiny$+1$})(k\mbox{\tiny$+1$})}$ from the main diagonal of $A$. Hence,

\[ \det(A) = a_{\mbox{\tiny$11$}} \det\left(A_{\mbox{\tiny$11$}}\right) = a_{\mbox{\tiny$11$}} \left(a_{\mbox{\tiny$22$}}a_{\mbox{\tiny$33$}} \cdots a_{(k\mbox{\tiny$+1$})(k\mbox{\tiny$+1$})} \right) = a_{\mbox{\tiny$11$}} a_{\mbox{\tiny$22$}}a_{\mbox{\tiny$33$}} \cdots a_{(k\mbox{\tiny$+1$})(k\mbox{\tiny$+1$})}\]

We have $\det(A)$ is the product of the entries along its main diagonal. This shows $P(k+1)$ is true, and, hence, by induction, the result holds for all $n \times n$ upper triangular matrices. The $n \times n$ identity matrix $I_{n}$ is a lower triangular matrix whose main diagonal consists of all $1$'s. Hence, $\det\left(I_{n}\right) = 1$, as required.

\end{enumerate}

\closegraphsfile

9.4: The Binomial Theorem

\newpage

\subsection{Exercises}

In Exercises \ref{simpfactfirst} - \ref{simpfactlast}, simplify the given expression.

\begin{multicols}{3}

\begin{enumerate}

\item $\left(3!\right)^2$ \label{simpfactfirst}

\item $\dfrac{10!}{7!}$

\item $\dfrac{7!}{2^3 3!}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\dfrac{9!}{4! 3! 2!}$

\item $\dfrac{(n+1)!}{n!}$, $n \geq 0$.

\item $\dfrac{(k-1)!}{(k+2)!}$, $k \geq 1$.

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\displaystyle{\binom{8}{3}}$

\item $\displaystyle{\binom{117}{0}}$

\item $\displaystyle{\binom{n}{n-2}}$, $n \geq 2$ \label{simpfactlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{pascalfirst} - \ref{pascallast}, use Pascal's Triangle to expand the given binomial.

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x+2)^5$ \label{pascalfirst}

\item $(2x-1)^4$

\item $\left(\frac{1}{3} x + y^2\right)^3$

\item $\left(x - x^{-1} \right)^{4}$ \label{pascallast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{pascalcomplexfirst} - \ref{pascalcomplexlast}, use Pascal's Triangle to simplify the given power of a complex number.

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(1+2i)^4$ \label{pascalcomplexfirst}

\item $\left(-1 + i \sqrt{3}\right)^3$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{2}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $\left(\dfrac{\sqrt{3}}{2} + \dfrac{1}{2}\, i\right)^3$

\item $\left(\dfrac{\sqrt{2}}{2} - \dfrac{\sqrt{2}}{2} \, i\right)^4$ \label{pascalcomplexlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

In Exercises \ref{usebinomfirst} - \ref{usenbinomlast}, use the Binomial Theorem to find the indicated term.

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item The term containing $x^3$ in the expansion $(2x-y)^{5}$ \label{usebinomfirst}

\item The term containing $x^{117}$ in the expansion $(x+2)^{118}$

\item The term containing $x^{\frac{7}{2}}$ in the expansion $\left(\sqrt{x}-3\right)^8$

\item The term containing $x^{-7}$ in the expansion $\left(2x - x^{-3} \right)^{5}$

\item The constant term in the expansion $\left(x + x^{-1} \right)^{8}$ \label{usenbinomlast}

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item Use the Prinicple of Mathematical Induction to prove $n! > 2^{n}$ for $n \geq 4$.

\item Prove $\displaystyle{\sum_{j=0}^{n} \binom{n}{j} = 2^{n}}$ for all natural numbers $n$. (HINT: Use the Binomial Theorem!)

\item With the help of your classmates, research \href{http://en.wikipedia.org/wiki/Pascal'...rline{Patterns and Properties of Pascal's Triangle}}.

\item You've just won three tickets to see the new film, `$8.\overline{9}$.' Five of your friends, Albert, Beth, Chuck, Dan, and Eugene, are interested in seeing it with you. With the help of your classmates, list all the possible ways to distribute your two extra tickets among your five friends. Now suppose you've come down with the flu. List all the different ways you can distribute the three tickets among these five friends. How does this compare with the first list you made? What does this have to do with the fact that $\binom{5}{2} = \binom{5}{3}$?

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\newpage

\subsection{Answers}

\begin{multicols}{3}

\begin{enumerate}

\item $36$

\item $720$

\item $105$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $1260$

\item $n+1$

\item $\frac{1}{k(k+1)(k+2)}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{3}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $56$

\item $1$

\item $\frac{n(n-1)}{2}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $(x+2)^5 = x^5+10x^4+40x^3+80x^2+80x+32$

\item $(2x-1)^4 = 16x^4-32x^3+24x^2-8x+1$

\item $\left(\frac{1}{3} x + y^2\right)^3 = \frac{1}{27} x^3+\frac{1}{3}x^2y^2+xy^4+y^6$

\item $\left(x - x^{-1} \right)^{4} = x^4-4x^2+6-4x^{-2}+x^{-4}$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\begin{multicols}{4}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $-7-24i$

\item $8$

\item $i$

\item $-1$

\setcounter{HW}{\value{enumi}}

\end{enumerate}

\end{multicols}

\begin{multicols}{5}

\begin{enumerate}

\setcounter{enumi}{\value{HW}}

\item $80x^3y^2$

\item $236x^{117}$

\item $-24x^{\frac{7}{2}}$

\item $-40 x^{-7}$

\item $70$

\end{enumerate}

\end{multicols}

\newpage

\closegraphsfile