Aenicillin, a type of bacteria, is grown in a petri dish. Its population doubles every six hours. Denoting the amount of Aenicillin at the beginning of the first day as $x_0$, (a) write down a recurrence relation for $x_n$, $n \ge 1$, the amount of Aenicillin in the petri dish at the end of $n^{\text{th}}$ day,
(b) find $x_n$, expressing your answer in the form of $x_n = \text{f}(n)$, $n \ge 1$. The growth rate of another bacteria, Benicillin, is known to follow the recurrence relation $u_n = u_{n-1} + n$, $n \ge 1$ where $u_n$ is the amount of Benicillin at the end of $n^{\text{th}}$ day. The amount of Benicillin in the petri dish at the beginning of the first day, $u_0$, is $1$ unit.

1. Determine the daily growth factor based on the doubling time

Since the population doubles every 6 hours, the number of doubling periods in a 24-hour day is $\frac{24}{6} = 4$.
Therefore, the population is multiplied by $2^4 = 16$ each day.

2. Formulate the recurrence relation connecting consecutive days

The amount at the end of the $n^{\text{th}}$ day ($x_n$) is 16 times the amount at the end of the $(n-1)^{\text{th}}$ day (which is exactly the beginning of the $n^{\text{th}}$ day, $x_{n-1}$).
$x_n = 16x_{n-1}$

1. Formulate the explicit formula for the $n^{\text{th}}$ term using the common ratio

From part (a), the recurrence relation is $x_n = 16x_{n-1}$.
This represents a geometric sequence with a common ratio of $16$.
Given the initial amount at $n = 0$ is $x_0$, the amount at the end of the $n^{\text{th}}$ day is given by:
$x_n = x_0(16)^n$

1. Find the values of $u_1 - u_0$, $u_2 - u_1$ and $u_3 - u_2$ using the given recurrence relation.

Given $u_n = u_{n-1} + n$, we have $u_n - u_{n-1} = n$.
For $n=1$, $u_1 - u_0 = 1$.
For $n=2$, $u_2 - u_1 = 2$.
For $n=3$, $u_3 - u_2 = 3$.

2. Express the sum in terms of the individual differences.

$(u_1 - u_0) + (u_2 - u_1) + \dots + (u_n - u_{n-1})$
$= 1 + 2 + \dots + n$

3. Find the sum of the resulting arithmetic series to obtain a quadratic expression.

This is an arithmetic series with the first term $a = 1$, common difference $d = 1$, and $n$ terms.
Sum $= \frac{n}{2}(a + l)$
$= \frac{n}{2}(1 + n)$
$= \frac{n(n+1)}{2}$ (or $\frac{1}{2}n^2 + \frac{1}{2}n$)

1. Expand the summation to show cancellation (method of differences).

$\sum_{r=1}^{n} (u_r - u_{r-1}) = (u_1 - u_0) + (u_2 - u_1) + (u_3 - u_2) + \dots + (u_n - u_{n-1})$
$= -u_0 + (u_1 - u_1) + (u_2 - u_2) + \dots + (u_{n-1} - u_{n-1}) + u_n$
$= u_n - u_0 \quad \text{(shown)}$

2. Calculate the sum of the differences using the result from part (c).

From the recurrence relation $u_r - u_{r-1} = r$, the sum from part (c) evaluates to an arithmetic progression:
$\sum_{r=1}^{n} (u_r - u_{r-1}) = 1 + 2 + 3 + \dots + n$
Using the sum formula for an arithmetic series with $a=1$, $d=1$ and $n$ terms:
$S_n = \frac{n}{2}[2(1) + (n-1)(1)] = \frac{n(n+1)}{2} = \frac{1}{2}n^2 + \frac{1}{2}n$

3. Equate the two expressions for the sum and solve for $u_n$.

Equating our results from the telescoping sum and the arithmetic progression:
$u_n - u_0 = \frac{1}{2}n^2 + \frac{1}{2}n$
Given the initial condition $u_0 = 1$:
$u_n - 1 = \frac{1}{2}n^2 + \frac{1}{2}n$
$u_n = \frac{1}{2}n^2 + \frac{1}{2}n + 1$