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\begin{document}

\module{The golden sequence}{02}{11}{2 periods}
\en

\prereq{Geometric sequences, the golden ratio}

\object{\begin{itemize}
        \item Use the properties of geometric sequences on an particularly interesting example.
	\item Discover some properties of the Golden Ratio.
	\end{itemize}}

\mater{\begin{itemize}
	\item Task sheet.
	\item Correction of the task sheet.
	\item Beamer (may not be needed).
	\end{itemize}}

\modpart{Pair work}{2 periods}

Working in pairs, students have to answer a series of questions about the golden sequence, the only sequence which is at the same time a Fibonacci sequence and a geometric sequence.



\pagebreak

\moddocdis{The golden sequence}{02}{11}{Task sheet}

The Golden number is the irrational real number defined as
$$\varphi=\frac{1+\sqrt{5}}{2}.$$
This famous number is strongly related to the Fibonacci sequence. It is also interesting to study the sequence defined as follows :
$$\left\{
\begin{array}{rcl}
b_1 &=& \frac{1}{\varphi} \\
b_{n+1} &=& b_n\times \varphi
\end{array}
\right.$$

\probpart{-- Working with approximate values}

\begin{enumerate}
\item What kind of sequence is $(b_n)$ ? Give the explicit definition of $b_n$.
\item Write the first seven terms of the sequence in the simplest way, using $\varphi$ and without doing any computation.
\item Give an approximate value to 3 DP of the first seven terms of the sequence.
\item Find the first value of $n$ for which $b_n>100$.
\item Check with the approximate values that for any integer $n$ lower than or equal to $5$, $$b_n+b_{n+1}=b_{n+2}.$$
\end{enumerate}


\probpart{-- Working with exact values.}

In this part, no approximate value must be used, so the calculator will be useless. All answers must be given in the form $a+b\sqrt{5}$ where $a$ and $b$ are rational numbers. For example, $\varphi$ will be written as
$$\varphi=\frac12+\frac12\sqrt{5}.$$

\begin{enumerate}
\item Check that $-\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right)=1$ and deduce the exact value of $b_1=\frac{1}{\varphi}$.
\item Compute  the first seven terms of the sequence $(b_n)$ and give the answers as explained at the beginning of this part.
\item Check, using the exact values, that for any integer $n$ lower than $5$, $b_n+b_{n+1}=b_{n+2}$.
\item Deduce the probable values of $b_8$, $b_9$, and $b_{10}$.
\item Check that $\frac{1}{\varphi}+1=\varphi$ and deduce that for any positive integer $n$, $$\varphi^{n}+\varphi^{n+1}=\varphi^{n+2}.$$
What property of the sequence $b$ does it prove ?
\end{enumerate}

\probpart{-- The sum of the first $n$ terms}

\begin{enumerate}
\item In a table, compare the approximate values (to 3DP) of the sums $b_1+b_2$, $b_1+b_2+b_3$ and $b_1+b_2+b_3+b_4$ with the values of $\varphi^2$, $\varphi^3$ and $\varphi^4$.
\item Deduce from the previous question a possible formula for $b_1+b_2+\ldots+b_n$.
\item Prove the formula.
\end{enumerate}

\pagebreak

\moddocdis{The golden sequence}{02}{11}{Answers}

\setcounter{probpart}{0}

\corrige

\probpart{-- Working with approximate values}

\begin{enumerate}
\item From its definition, we can say that $(b_n)$ is a geometric sequence with first term $\frac{1}{\varphi}$ and common ratio $\varphi$. The explicit definition of $b_n$ is therefore $$b_n=\frac{1}{\varphi}\times \varphi^{n-1}=\varphi^{n-2}.$$
\item Using the explicit formula, we get $b_1=\frac{1}{\varphi}$, $b_2=\varphi^0=1$, $b_3=\varphi^1=\varphi$, $b_4=\varphi^2$, $b_5=\varphi^3$, $b_6=\varphi^4$, $b_7=\varphi^5$.
\item Using the previous answers and a calculator, we find $b_1\simeq0.618$, $b_2=1$, $b_3\simeq1.618$, $b_4\simeq2.618$, $b_5\simeq4.236$, $b_6\simeq6.854$, $b_7\simeq11.090$.
\item The first value of $n$ such that $b_n>100$ is $n=12$.
\item For $n=1$, $b_1+b_2=0.618+1=1.618=b_3$.\\ 
For $n=2$, $b_2+b_3=1+1.618=2.618=b_4$.\\ 
For $n=3$, $b_3+b_4=1.618+2.618=4.236=b_5$.\\
For $n=4$, $b_4+b_5=2.618+4.236=6.854=b_6$.\\
For $n=5$, $b_5+b_6=4.236+6.854=11.090=b_7$.
\end{enumerate}


\probpart{-- Working with exact values.}


\begin{enumerate}
\item \begin{eqnarray*}
      -\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(\frac{1}{2}+\frac{\sqrt{5}}{2}\right) &=& -\left(\left(\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2\right) \\
	&=& -\left(\frac{1}{4}-\frac{5}{4}\right) \\
	&=& -\left(-\frac{4}{4}\right) \\
	&=& 1
      \end{eqnarray*}
This means that $-\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)\times\varphi=1$, so $$b_1=\frac{1}{\varphi}=-\left(\frac{1}{2}-\frac{\sqrt{5}}{2}\right)=-\frac{1}{2}+\frac{1}{2}\sqrt{5}.$$
\item As we've seen previously, $b_1=-\frac{1}{2}+\frac{1}{2}\sqrt{5}$. Also, it's obvious that $b_2=1=1+0\sqrt{5}$ and we know that $b_3=\varphi=\frac{1}{2}+\frac{1}{2}\sqrt{5}$.Let's look at the next terms.\\
$b_4=\varphi^2=\left(\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)^2=\frac14+\frac{1}{2}\sqrt{5}+\frac54=\frac32+\frac{1}{2}\sqrt{5}$.\\
$b_5=\varphi^3=\varphi^2\times \varphi= \left(\frac32+\frac{1}{2}\sqrt{5}\right)\times\left(\frac{1}{2}+\frac{1}{2}\sqrt{5}\right)
=\frac34+\frac34\sqrt{5}+\frac{1}{4}\sqrt{5}+\frac{5}{4}=2+1\sqrt{5}$.\\
$b_6=\varphi^4={(\varphi^2)}^2=\left(\frac32+\frac{1}{2}\sqrt{5}\right)^2=\frac94+\frac{3}{2}\sqrt{5}+\frac54=\frac72+\frac{3}{2}\sqrt{5}$.\\
$b_7=\varphi^5=\varphi^3\times \varphi^2= \left(2+1\sqrt{5}\right)\times\left(\frac32+\frac{1}{2}\sqrt{5}\right)
=3+\sqrt{5}+\frac{3}{2}\sqrt{5}+\frac{5}{2}=\frac{11}{2}+\frac{5}{2}\sqrt{5}$.
\item For $n=1$, $b_1+b_2=-\frac{1}{2}+\frac{1}{2}\sqrt{5}+1+0\sqrt5=\frac{1}{2}+\frac{1}{2}\sqrt{5}=b_3$.\\ 
For $n=2$, $b_2+b_3=1+0\sqrt5+\frac{1}{2}+\frac{1}{2}\sqrt{5}=\frac32+\frac{1}{2}\sqrt{5}=b_4$.\\ 
For $n=3$, $b_3+b_4=\frac{1}{2}+\frac{1}{2}\sqrt{5}+\frac32+\frac{1}{2}\sqrt{5}=2+1\sqrt{5}=b_5$.\\
For $n=4$, $b_4+b_5=\frac32+\frac{1}{2}\sqrt{5}+2+1\sqrt{5}=\frac72+\frac{3}{2}\sqrt{5}=b_6$.\\
For $n=5$, $b_5+b_6=2+1\sqrt{5}+\frac72+\frac{3}{2}\sqrt{5}=\frac{11}{2}+\frac{5}{2}\sqrt{5}=b_7$.
\item We can expect that :\\
$b_8=b_6+b_7=\frac72+\frac{3}{2}\sqrt{5}+\frac{11}{2}+\frac{5}{2}\sqrt{5}=9+4\sqrt{5}$\\
$b_9=b_7+b_8=\frac{11}{2}+\frac{5}{2}\sqrt{5}+9+4\sqrt{5}=\frac{29}{2}+\frac{13}{2}\sqrt{5}$\\
$b_{10}=b_8+b_9=9+4\sqrt{5}+\frac{29}{2}+\frac{13}{2}\sqrt{5}=\frac{47}{2}+\frac{21}{2}\sqrt{5}$.
\item We proved in the first question that $\frac{1}{\varphi}=-\frac{1}{2}+\frac{1}{2}\sqrt{5}$, so $$\frac{1}{\varphi}+1=-\frac{1}{2}+\frac{1}{2}\sqrt{5}+1=\frac{1}{2}+\frac{1}{2}\sqrt{5}=\varphi.$$
Multiplying this equation by $\varphi^{n+1}$, we deduce that for any positive integer $n$, 
\begin{eqnarray*}
\varphi^{n+1}\times\left(\frac{1}{\varphi}+1\right) &=& \varphi^{n+1}\times \varphi \\
\frac{\varphi^{n+1}}{\varphi}+\varphi^{n+1} &=& \varphi^{n+1}\times \varphi \\
\varphi^{n}+\varphi^{n+1} &=& \varphi^{n+2}
\end{eqnarray*}
Using the explicit formula of $b_n$, this proves that for any positive integer $n$, $b_{n+2}+b_{n+3}=b_{n+4}$ and so, as its also true for $b_1$ and $b_2$, 
$$b_{n}+b_{n+1}=b_{n+2}.$$
\end{enumerate}

\probpart{-- The sum of the first $n$ terms}

\begin{enumerate}
\item Comparison between the two sequences :
\begin{center}
\begin{tabular}{|c|c|c|c|}
\hline
$b_1+b_2$ & $1.618$ & $\varphi^2$ & $2.618$\\
\hline
$b_1+b_2+b_3$ & $3.236$ & $\varphi^3$ & $4.236$\\
\hline
$b_1+b_2+b_3+b_4$ & $5.854$ & $\varphi^4$ & $6.854$ \\
\hline
\end{tabular}
\end{center}
\item From the results in the previous table, it seems that $b_1+b_2+\ldots+b_n=\varphi^n-1$.
\item Using the formula for the sum of the terms of a geometric sequence, we get
\begin{eqnarray*}
b_1+b_2+\ldots+b_n &=& \frac{1}{\varphi}\times\frac{1-\varphi^n}{1-\varphi}
\end{eqnarray*}
As we know that $\frac{1}{\varphi}+1=\varphi$, we can say that $\frac{1}{\varphi}=\varphi-1$, and so 
\begin{eqnarray*}
b_1+b_2+\ldots+b_n &=& (\varphi-1)\times\frac{1-\varphi^n}{1-\varphi}\\
b_1+b_2+\ldots+b_n &=& -(1-\varphi)\times\frac{1-\varphi^n}{1-\varphi}\\
b_1+b_2+\ldots+b_n &=& -(1-\varphi^n)\\
b_1+b_2+\ldots+b_n &=& \varphi^n-1
\end{eqnarray*}
\end{enumerate}



\end{document}
