(Candidates who have followed the specialization course)We observe the size of an ant colony every day. For any non-zero natural number $n$, we denote by $u _ { n }$ the number of ants, expressed in thousands, in this population at the end of the $n$-th day. At the beginning of the study the colony has 5000 ants and after one day it has 5100 ants. Thus, we have $u _ { 0 } = 5$ and $u _ { 1 } = 5.1$. We assume that the increase in the size of the colony from one day to the next decreases by $10\%$ each day. In other words, for any natural number $n$,
$$u _ { n + 2 } - u _ { n + 1 } = 0.9 \left( u _ { n + 1 } - u _ { n } \right) .$$
- Prove that under these conditions, $u _ { 2 } = 5.19$.
- For any natural number $n$, we set $V _ { n } = \binom { u _ { n + 1 } } { u _ { n } }$ and $A = \left( \begin{array} { c c } 1.9 & - 0.9 \\ 1 & 0 \end{array} \right)$. a. Prove that, for any natural number $n$, we have $V _ { n + 1 } = A V _ { n }$.
We then admit that, for any natural number $n$, $V _ { n } = A ^ { n } V _ { 0 }$. b. We set $P = \left( \begin{array} { c c } 0.9 & 1 \\ 1 & 1 \end{array} \right)$. We admit that the matrix $P$ is invertible.
Using a calculator, determine the matrix $P ^ { - 1 }$. By detailing the calculations, determine the matrix $D$ defined by $D = P ^ { - 1 } A P$. c. Prove by induction that, for any natural number $n$, we have $A ^ { n } = P D ^ { n } P ^ { - 1 }$. For any natural number $n$, we admit that
$$A ^ { n } = \left( \begin{array} { c c }
- 10 \times 0.9 ^ { n + 1 } + 10 & 10 \times 0.9 ^ { n + 1 } - 9 \\
- 10 \times 0.9 ^ { n } + 10 & 10 \times 0.9 ^ { n } - 9
\end{array} \right) .$$
d. Deduce that, for any natural number $n$, $u _ { n } = 6 - 0.9 ^ { n }$.
\setcounter{enumi}{2} - Calculate the size of the colony at the end of the $10 ^ { \mathrm { th } }$ day. Round the result to the nearest ant.
- Calculate the limit of the sequence $(u _ { n })$. Interpret this result in context.