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\module{Geometric sequences}{02}{09}{3 periods}
\en
\nocouleur

\prereq{}

\object{\begin{itemize}
        \item Discover the concept of geometric sequence.
	\item Use the main formulae about geometric sequences.
	\end{itemize}}

\mater{\begin{itemize}
	\item Lesson about geometric sequences.
	\item Exercises about geometric sequences.
	\item Beamer
	\end{itemize}}

\modpart{Lesson}{30 mins}

The main results about geometric sequences are shown with a beamer.

\modpart{Exercises}{Remaining time}

Exercises about geometric sequences have to be done in groups of 3 or 4 students.

\pagebreak


\moddocdis{Geometric sequences}{02}{09}{Lesson}

\section{Definition and criterion}

\txt{A sequence of numbers $(b_n)$ is geometric if the quotient
between two consecutive terms is a constant number. Intuitively, to
go from one term to the next one, we always multiply by the same number.}

\defi{Geometric sequence}{A sequence of numbers $(b_n)$ is geometric if, for any
positive integer $n$, $\frac{b_{n+1}}{b_n}=q$ where $q$ is a fixed
real number, called the \emph{common ratio} of the sequence. We
can also write that $b_{n+1}=b_n\times q$. This equality is called
the \emph{recurrence relation} of the sequence.}

\section{Relations between terms}

\prop{Explicit definition}{For any positive integer $n$, $b_n=b_1\times q^{n-1}$. This
equality is called the \emph{explicit definition} of the sequence.}

\dem{First, this equality is true when $n=0$, as $b_1=b_1\times
q^0=b_1\times q^{1-1}$. Then, suppose that it is true for a value
$n=k$, meaning that $b_k=b_1\times q^{k-1}$. Then, from the
definition of the sequence, $b_{k+1}=b_k\times q=b_1\times
q^{k-1}\times q=b_1+\times q^k$. So the formula is true for $n=k+1$
too. So it's true for $n=0$, $n=1$, $n=2$, $n=3$, etc, for all
values of $n$.}

\prop{Relation between two terms}{For any two positive integers $n$ and $m$, $b_n=b_m\times
q^{n-m}$.}

\dem{From the explicit definition of the sequence $(b_n)$,
$b_n=b_1\times q^{n-1}$ and $b_m=b_1\times q^{m-1}$, so
$\frac{b_n}{b_m}=\frac{b_1\times q^{n-1}}{b_m=b_1\times
q^{m-1}}=\frac{q^{n-1}}{q^{m-1}}=q^{n-m}$. Therefore, $b_n=b_m\times
q^{n-m}$.}

\pagebreak

\section{Limit when $n$ approaches $+\infty$}

\theo{Limit of a geometric sequence}{The limit of a geometric sequence $(b_n)$ of common ratio
$q$ and first term~$b_1$
\begin{itemize}
\itemb is equal to $0$, trivially, if $b_1=0$ ;
\itemb is equal to $b_1$, trivially, if $r=1$ ;
\itemb is equal to $+\infty$ when $b_1>0$ and $q>1$ ;
\itemb is equal to $-\infty$ when $b_1<0$ and $q>1$ ;
\itemb is equal to $0$ if $q\in]-1;1[$ ;
\itemb doesn't exist when $r\leq -1$.
\end{itemize}
In the last situation, the sequence is said to be \emph{divergent}.}

%\proo{\begin{itemize}
%\itemb If $b_1=0$ or if $r=1$, the result is obvious : in both
%cases, the sequence is constant !
%\itemb If $b_1>0$ and $q>1$, consider any real number $K$. The
%inequation $b_n>K$, or $b_1\times q^{n-1}>K$ is equivalent to
%\begin{eqnarray*}
%q^{n-1} &>& \frac{K}{b_1}\\
%\ln(q^{n-1}) &>& \ln\left(\frac{K}{b_1}\right)\\
%(n-1)\ln q &>& \ln K-\ln b_1\\
%(n-1) &>& \frac{\ln K-\ln b_1}{\ln q}\\
%n &>& \frac{\ln K-\ln b_1}{\ln q}+1
%\end{eqnarray*}
%This means that for any real number $K$, there exist some integer
%$N$ such that for any $n\geq N$, $b_N>K$. This is exactly the
%definition of the fact that $\lim b_n=+\infty$.
%\itemb If $b_1<0$ and $q>1$, consider any real number $K$. The
%inequation $b_n<K$, or $b_1\times q^{n-1}<K$ is equivalent to
%\begin{eqnarray*}
%q^{n-1} &>& \frac{K}{b_1}\\
%\ln(q^{n-1}) &>& \ln\left(\frac{K}{b_1}\right)\\
%(n-1)\ln q &>& \ln K-\ln b_1\\
%(n-1) &>& \frac{\ln K-\ln b_1}{\ln q}\\
%n &>& \frac{\ln K-\ln b_1}{\ln q}+1
%\end{eqnarray*}
%This means that for any real number $K$, there exist some integer
%$N$ such that for any $n\geq N$, $b_N<K$. This is exactly the
%definition of the fact that $\lim b_n=-\infty$.
%\itemb If $q\in]-1;1[$, consider any positive number $\varepsilon$. The
%inequation $|b_n|<\varepsilon$, or $|b_1|\times
%|q|^{n-1}<\varepsilon$ is equivalent to
%\begin{eqnarray*}
%|q|^{n-1} &<& \frac{\varepsilon}{|b_1|}\\
%\ln(|q|^{n-1}) &<& \ln\left(\frac{\varepsilon}{|b_1|}\right)\\
%(n-1)\ln |q| &<& \ln \varepsilon-\ln |b_1|\\
%(n-1) &>& \frac{\ln \varepsilon-\ln |b_1|}{\ln |q|}\\
%n &>& \frac{\ln \varepsilon-\ln |b_1|}{\ln |q|}+1
%\end{eqnarray*}
%This means that for any positive real number $\varepsilon$, there
%exists some integer $N$ such that for any $n\geq N$,
%$|b_N|<\varepsilon$. This is exactly the definition of the fact that
%$\lim b_n=0$.
%\itemb Finally when $r\leq -1$, the sequence is alternating between
%positive and negative terms, whose absolute values approach
%$+\infty$. So the sequence has no limit.
%\end{itemize}
%%
%
%}



\section{Sums of consecutive terms}

\theo{Sum of consecutive terms}{Let $(b_n)$ be an geometric sequence, The sum $S$ of the $n$ first consecutive terms,
defined as
$S=b_1+b_2+\ldots+b_{n-1}+b_n$, or more precisely
$S=\sum_{i=1}^{n} b_i$, is given by the formula
$$S=b_1\frac{1-q^{n}}{1-q}.$$}

\dem{Using the formula of proposition 2.2, we can write $S$ as
\begin{eqnarray*}
S &=& b_1+b_1\times q+\ldots+b_1\times q^{n-2}+b_1\times q^{n-1}\\
  &=& b_1\times(1+q+\ldots+q^{n-2}+q^{n-1}).
\end{eqnarray*}
But, by a simple expansion, we see that
$(1+q+\ldots+q^{n-2}+q^{n-1})(1-q)=1-q^{n}$ and so
$$S=b_1\frac{1-q^{n}}{1-q}.$$}






\pagebreak

\moddocdis{Geometric sequences}{02}{09}{Exercises}

\newcounter{chap}
\setcounter{chap}{9}

\exoc{You complain that the hot tub in your hotel suite is not hot enough. The hotel tells you that they will increase
the temperature by 10\%\ each hour.  If the current temperature of the hot tub is 75$^\circ$ F, what will be the
temperature of the hot tub after 3 hours, to the nearest tenth of a degree?}

\exoc{A culture of bacteria doubles every 2 hours.  If there are 500 bacteria at the beginning, how many bacteria will
there be after 24 hours?}

\exoc{A mine worker discovers an ore sample containing 500 mg of radioactive material.  It is discovered that the
radioactive material has a half life of 1 day.  Find the amount of radioactive material in the sample at the beginning
of the 7th day.}

\exoc{Here are the first three levels in a school group telephoning tree.

\begin{center}
\begin{tabular}{cccc}
\multicolumn{4}{c}{Teacher}\\
\multicolumn{2}{c}{Student 1} & \multicolumn{2}{c}{Student 2}\\
Student 3 & Student 4 & Student 5 & Student 6 \\
\end{tabular}
\end{center}

At what level are all 53 students in the group contacted?}

\exoc{Consider the geometric sequences $d$ and $d$ starting like
this :
$b=\{\frac{1}{2},\frac{3}{2},\frac{9}{2},\frac{27}{2},\frac{81}{2},\ldots\}$
and
$d=\{\frac{7}{3},\frac{7}{6},\frac{7}{12},\frac{7}{24},\frac{7}{48},\ldots\}$.
\begin{enumerate}
\item Compute the common ratio $q$ for each sequence.
\item Find recursive definitions for these two geometric sequences.
\item Find the formula of the $n$-th term for each sequence.
\item Compute the $15$-th term of each sequence.
\end{enumerate}
}


\exoc{Bill and Steve decide to buy the same computer. They don't have the money at their disposal yet, so the seller offers two types of credits. In each case, they have to pay a certain amount at the time of the actual sell, then the remainder of the 2000 euros  in monthly instalments.\\
Bill chooses to pay $80$ euros first, then monthly instalments of 160 euros for each of the 12 following months. Steve chooses to pay $125$ euros first, and the following instalments with a monthly increase of 3\,\% over the 11 following months. On the twelfth month, he pays whatever is left of the 2000 euros.

\probpart{-- Bill's choice}

Let's note $u_{1}$ the initial amount paid by Bill and $u_{n}$ the total amount paid after $n$ month. So, $u_{1} = 80$ and $u_{2}$ is the total amount he has paid at the end of the first month.
\begin{enumerate}
\item Compute $u_{2}$ and $u_{3}$.
\item   \begin{enumerate}
\item  What kind of sequence is $\left(u_{n}\right)$ ? Explain.
\item  Write $u_{n}$ as a function of $n$.
\end{enumerate}
\end{enumerate}

\probpart{--  Steve's choice}

Let's note $v_{1}$ the initial amount paid by Steve and $v_{n}$ the total amount paid on the $n$-th month (with $n$ between $2$ and $11$), rounded to the closest integer. So, $v_{1} = 125$ and $v_{2} = 129$.

\begin{enumerate}
\item Compute the value of $v_{2}$ rounded to the closest integer.
\item What kind of sequence is $\left(v_{n}\right)$ ? Explain.
\item What is the total amount paid by Steve at the end of the $11$-th month ? What must he pay for the $12$-th month ?
\item From what moment are Steve's monthly instalments greater then Bill's ?
\end{enumerate}}


\exoc{Write the formula for the $n$-th term of the geometric
sequences with the following :
\begin{enumerate}
\item $u_5=18750$ and $q=5$ ;
\item $u_5=3$ and $q=\frac12$ ;
\item $u_8=472392$ and $q=3$ ;
\item $u_5=15625$ and $q=-5$ ;
\item $u_{11}=20480$ and $u_8=5120$ ;
\item $u_{9}=98415$ and $u_{12}=2657205$ ;
\item $u_{5}=54432$ and $u_8=11757312$.
\end{enumerate}
}




\exoc{Let $(c_n)$ be the sequence defined by
$$\left\{
\begin{array}{rcl}
c_1 & = & \dis\frac{1}{3} \\
c_{n+1} & = & \dis\frac{n+1}{3n}c_n \\
\end{array}
\right.
$$
\begin{enumerate}
\item Compute the terms from  $c_1$ to $c_6$.
\item Let $(b_n)$ be the sequence defined by $$b_n=\frac{c_n}{n}.$$
Prove that $(b_n)$ is geometric and give its common ratio.
\item Give the explicit formula of $(b_n)$ and deduce the one of $(c_n)$.
\item Compute the sum $\sum_{k=1}^{20}b_k.$
\item Determine the limit of the sequence $(b_n)$.
\item Admitting that for any natural number $n$, $n<2^n$, prove that $$0\leq
c_n<\left(\frac{2}{3}\right)^n.$$
\item Deduce from the previous question the limit of the sequence $(c_n)$.
\end{enumerate}}



\end{document}
