Anand writes a computer programme to simulate a population of organisms in a controlled environment. It is assumed that none of the organisms die or leave the environment within the duration of a simulation.

1. Formulate an expression for the number of organisms at the end of Day $n$.

The number of organisms forms an arithmetic sequence with first term $a = 200$ and common difference $d = 48$.
The number of organisms at the end of Day $n$ is $200 + 48(n - 1)$.

2. Set up and solve an inequality to find the first day the population exceeds 2025.

$200 + 48(n - 1) > 2025$
$48(n - 1) > 1825$
$n - 1 > \frac{1825}{48}$
$n - 1 > 38.0208...$
$n > 39.0208...$
Since $n$ must be an integer, the smallest possible value is $n = 40$.

1. Identify the pattern as a geometric progression and find the common ratio.

Let $u_n$ be the number of organisms at the end of Day $n$.
Given $u_1 = 15$.
For each subsequent day, each organism spawns 2 more, so the total becomes the original 1 plus the 2 new ones. Thus, the population is multiplied by 3 each day.
$u_2 = 15 + 15(2) = 45 = 15 \times 3$
This forms a geometric progression with first term $a = 15$ and common ratio $r = 3$.

2. Use the $n$th term formula for a geometric progression to find the population at the end of Day 20.

Using the $n$th term formula $u_n = ar^{n-1}$ with $n = 20$:
$u_{20} = 15(3)^{20-1}$
$u_{20} = 15(3)^{19}$
$u_{20} = 17\,433\,922\,005$

1. Determine the multipliers for each process and formulate an equation for the final population.

In Process I, each organism spawns 3 more, so the population is multiplied by $1 + 3 = 4$.
In Process II, each organism spawns 5 more, so the population is multiplied by $1 + 5 = 6$.
From Day 2 to Day 9, the processes occur for $9 - 2 + 1 = 8$ days in total.
The population undergoes Process I for $m$ days and Process II for $(8 - m)$ days.
Equation for the population at the end of Day 9:
$5 \times 4^m \times 6^{8-m} = 1105920$

2. Solve the equation for $m$ using laws of indices.

$4^m \times \frac{6^8}{6^m} = \frac{1105920}{5}$
$\left(\frac{4}{6}\right)^m \times 1679616 = 221184$
$\left(\frac{2}{3}\right)^m = \frac{221184}{1679616}$
$\left(\frac{2}{3}\right)^m = \frac{32}{243}$
Since $\frac{32}{243} = \frac{2^5}{3^5}$:
$\left(\frac{2}{3}\right)^m = \left(\frac{2}{3}\right)^5$
$m = 5$
(Note: Students can also solve this by systematically testing integer values for $m$ from 0 to 8).

1. Express the population size for the first few days to find a pattern.

Let $P_n$ be the population at the start of Day $n$, after the introduction of new organisms.
$P_1 = 100$
$P_2 = 100(0.9) + 20$
$P_3 = [100(0.9) + 20](0.9) + 20 = 100(0.9)^2 + 20(0.9) + 20$
$P_4 = [100(0.9)^2 + 20(0.9) + 20](0.9) + 20 = 100(0.9)^3 + 20(0.9)^2 + 20(0.9) + 20$

2. Generalise the pattern for $P_n$ and identify the geometric series.

Following the pattern, for Day $n$:
$P_n = 100(0.9)^{n-1} + \left[ 20(0.9)^{n-2} + 20(0.9)^{n-3} + \dots + 20(0.9) + 20 \right]$
The terms in the square brackets form a geometric progression with first term $a=20$, common ratio $r=0.9$, and the number of terms is $n-1$.

3. Apply the sum formula for a geometric progression to simplify the expression.

$S_{n-1} = \frac{a(1-r^{n-1})}{1-r} = \frac{20(1-0.9^{n-1})}{1-0.9}$
$S_{n-1} = \frac{20(1-0.9^{n-1})}{0.1} = 200(1-0.9^{n-1})$

4. Combine the terms to find the final expression for $P_n$ in the required form.

$P_n = 100(0.9)^{n-1} + 200(1-0.9^{n-1})$
$P_n = 100(0.9)^{n-1} + 200 - 200(0.9)^{n-1}$
$P_n = 200 - 100(0.9)^{n-1}$
This is in the form $s - t(r^{n-1})$ where $s=200$, $t=100$, and $r=0.9$.

1. Determine the expression for $P_n$ from part (d)(i) to use for the long-term behaviour.

Let $P_n$ be the population at the start of Day $n$.
$P_1 = 100$
$P_2 = 100(0.9) + 20$
Following the pattern,
$P_n = 100(0.9)^{n-1} + 20(0.9)^{n-2} + \dots + 20$
Using the sum of a geometric progression for the terms involving $20$,
$P_n = 100(0.9)^{n-1} + \frac{20(1 - 0.9^{n-1})}{1 - 0.9}$
$P_n = 100(0.9)^{n-1} + 200(1 - 0.9^{n-1}) = 200 - 100(0.9)^{n-1}$

2. Evaluate the limit of $P_n$ as $n$ approaches infinity.

In the long term, $n \to \infty$.
Since $0 < 0.9 < 1$, $(0.9)^{n-1} \to 0$ as $n \to \infty$.
Therefore, $P_n \to 200 - 100(0) = 200$.

1. Explain that the term containing the initial population diminishes to zero in the long term.

If the initial population on Day 1 is $A$, the population on Day $n$ generally takes the form $P_n = 200 + (A - 200)(0.9^{n-1})$.
As $n \to \infty$, since the common ratio is $0.9$ (which is strictly between $-1$ and $1$), $0.9^{n-1} \to 0$.
Therefore, the term containing the initial population $A$ vanishes as $n$ becomes very large, meaning the long-term population size will always tend to 200 regardless of $A$.