Exercise 4 — Candidates who have followed the specialization course A fish farmer has two basins A and B for raising his fish. Every year at the same time:
he empties basin B and sells all the fish it contained and transfers all the fish from basin A to basin B;
the sale of each fish allows the purchase of two small fish intended for basin A.
Furthermore, the fish farmer buys an additional 200 fish for basin A and 100 fish for basin B. For every natural integer greater than or equal to 1, we denote respectively by $a _ { n }$ and $b _ { n }$ the numbers of fish in basins A and B after $n$ years. At the beginning of the first year, the number of fish in basin A is $a _ { 0 } = 200$ and that in basin B is $b _ { 0 } = 100$.
Justify that $a _ { 1 } = 400$ and $b _ { 1 } = 300$ then calculate $a _ { 2 }$ and $b _ { 2 }$.
We denote by $A$ and $B$ the matrices such that $A = \left( \begin{array} { l l } 0 & 2 \\ 1 & 0 \end{array} \right)$ and $B = \binom { 200 } { 100 }$ and for every natural integer $n$, we set $X _ { n } = \binom { a _ { n } } { b _ { n } }$. a. Explain why for every natural integer $n , X _ { n + 1 } = A X _ { n } + B$. b. Determine the real numbers $x$ and $y$ such that $\binom { x } { y } = A \binom { x } { y } + B$. c. For every natural integer $n$, we set $Y _ { n } = \binom { a _ { n } + 400 } { b _ { n } + 300 }$. Prove that for every natural integer $n , Y _ { n + 1 } = A Y _ { n }$.
For every natural integer $n$, we set $Z _ { n } = Y _ { 2 n }$. a. Prove that for every natural integer $n , Z _ { n + 1 } = A ^ { 2 } Z _ { n }$. Deduce that for every natural integer $n , Z _ { n + 1 } = 2 Z _ { n }$. b. We admit that this recurrence relation allows us to conclude that for every natural integer $n$, $$Y _ { 2 n } = 2 ^ { n } Y _ { 0 }$$ Deduce that $Y _ { 2 n + 1 } = 2 ^ { n } Y _ { 1 }$ then prove that for every natural integer $n$, $$a _ { 2 n } = 600 \times 2 ^ { n } - 400 \text { and } a _ { 2 n + 1 } = 800 \times 2 ^ { n } - 400 .$$
Basin A has a capacity limited to 10000 fish. a. An algorithm is given that, for a given value of $p$, computes the number of fish $a$ in basin A after $p$ years using the formulas $a_{2n} = 600 \times 2^n - 400$ and $a_{2n+1} = 800 \times 2^n - 400$. Use this algorithm to determine from which year the capacity of basin A is exceeded.
\textbf{Exercise 4 — Candidates who have followed the specialization course}
A fish farmer has two basins A and B for raising his fish. Every year at the same time:
\begin{itemize}
\item he empties basin B and sells all the fish it contained and transfers all the fish from basin A to basin B;
\item the sale of each fish allows the purchase of two small fish intended for basin A.
\end{itemize}
Furthermore, the fish farmer buys an additional 200 fish for basin A and 100 fish for basin B.\\
For every natural integer greater than or equal to 1, we denote respectively by $a _ { n }$ and $b _ { n }$ the numbers of fish in basins A and B after $n$ years.\\
At the beginning of the first year, the number of fish in basin A is $a _ { 0 } = 200$ and that in basin B is $b _ { 0 } = 100$.
\begin{enumerate}
\item Justify that $a _ { 1 } = 400$ and $b _ { 1 } = 300$ then calculate $a _ { 2 }$ and $b _ { 2 }$.
\item We denote by $A$ and $B$ the matrices such that $A = \left( \begin{array} { l l } 0 & 2 \\ 1 & 0 \end{array} \right)$ and $B = \binom { 200 } { 100 }$ and for every natural integer $n$, we set $X _ { n } = \binom { a _ { n } } { b _ { n } }$.\\
a. Explain why for every natural integer $n , X _ { n + 1 } = A X _ { n } + B$.\\
b. Determine the real numbers $x$ and $y$ such that $\binom { x } { y } = A \binom { x } { y } + B$.\\
c. For every natural integer $n$, we set $Y _ { n } = \binom { a _ { n } + 400 } { b _ { n } + 300 }$.\\
Prove that for every natural integer $n , Y _ { n + 1 } = A Y _ { n }$.\\
\item For every natural integer $n$, we set $Z _ { n } = Y _ { 2 n }$.\\
a. Prove that for every natural integer $n , Z _ { n + 1 } = A ^ { 2 } Z _ { n }$. Deduce that for every natural integer $n , Z _ { n + 1 } = 2 Z _ { n }$.\\
b. We admit that this recurrence relation allows us to conclude that for every natural integer $n$,
$$Y _ { 2 n } = 2 ^ { n } Y _ { 0 }$$
Deduce that $Y _ { 2 n + 1 } = 2 ^ { n } Y _ { 1 }$ then prove that for every natural integer $n$,
$$a _ { 2 n } = 600 \times 2 ^ { n } - 400 \text { and } a _ { 2 n + 1 } = 800 \times 2 ^ { n } - 400 .$$
\item Basin A has a capacity limited to 10000 fish.\\
a. An algorithm is given that, for a given value of $p$, computes the number of fish $a$ in basin A after $p$ years using the formulas $a_{2n} = 600 \times 2^n - 400$ and $a_{2n+1} = 800 \times 2^n - 400$. Use this algorithm to determine from which year the capacity of basin A is exceeded.
\end{enumerate}