The reproduction number, $R_0$, of a disease is a mathematical term that indicates how contagious an infectious disease is. It gives the average number of people who will contract a contagious disease from a person with that disease. That replication will continue if no one has been vaccinated against the disease or is already immune to it in his or her community. The infectious period of a disease is the time period during which an infected person can transmit the disease to any susceptible person he/she contacts. For example, if a disease has an $R_0$ of 3, a person who has the disease will transmit it to an average of 3 other people during the infectious period. Given that the infectious period is 2 weeks and initially $x$ people were infected, the average number of people infected is $3x$ in week 2, $3^2x$ in week 4, $3^3x$ in week 6, and so on. In a large population of people with no immunity to a virus, initially one person was infected with the virus. It is given that this virus has an $R_0$ of 2.5 and the infectious period is 2 weeks.

1. Formulate the inequality for the number of infected people in the $k$-th infectious period.

Let $k$ be the number of 2-week infectious periods.
Based on the given model, the number of newly infected people in period $k$ (which corresponds to week $2k$) forms a geometric sequence:
$u_k = 1 \times 2.5^k$
We require the number of infected people in a cycle to be at least $1000$:
$2.5^k \ge 1000$

2. Solve the exponential inequality to find $k$, and hence find the least number of weeks.

Taking the natural logarithm on both sides:
$k \ln 2.5 \ge \ln 1000$
$k \ge \frac{\ln 1000}{\ln 2.5}$
$k \ge 7.538$
Since $k$ represents discrete infectious periods, it must be an integer. The least integer value for $k$ is $8$.
Thus, the least number of weeks is $2k = 2(8) = 16$. (Shown)

1. Set up the inequality for the total reaching 2000

After week $2n$ the number infected is $2.5^{n}$ (initial 1 person, common ratio $2.5$). To infect the next 1000 people, the total infected must reach $1000+1000 = 2000$:
$2.5^{n} \ge 2000.$

2. Solve using logarithms and convert to weeks

Taking logarithms:
$n \ge \frac{\lg 2000}{\lg 2.5} = \frac{3.30103}{0.39794} = 8.30.$
Since $n$ must be an integer, the least value is $n = 9$.
Check: $2.5^{8} = 1526 < 2000$ and $2.5^{9} = 3815 \ge 2000$. ✓
Number of weeks $= 2n = 2(9) = 18$.

1. Set up equations by substituting values from $1$ to $n$ and sum them to obtain $u_n - u_0$.

Listing the equations for consecutive values of the index:
$u_1 - u_0 = 2^1 + 1 - 1$
$u_2 - u_1 = 2^2 + 2 - 1$
$u_3 - u_2 = 2^3 + 3 - 1$
$\vdots$
$u_n - u_{n-1} = 2^n + n - 1$

Summing all $n$ equations vertically, the terms on the left-hand side telescope (cancel out) to leave only $u_n - u_0$:
$u_n - u_0 = (2^1 + 2^2 + \dots + 2^n) + (0 + 1 + 2 + \dots + n - 1)$

2. Evaluate the sum of the geometric progression part.

The first group of terms is a geometric progression with first term $a = 2$, common ratio $r = 2$, and $n$ terms:
$S_{GP} = \frac{a(r^n - 1)}{r - 1}$
$S_{GP} = \frac{2(2^n - 1)}{2 - 1} = 2^{n+1} - 2$

3. Evaluate the sum of the arithmetic progression part.

The second group of terms is an arithmetic progression with first term $a = 0$, last term $l = n-1$, and $n$ terms:
$S_{AP} = \frac{n}{2}(a + l)$
$S_{AP} = \frac{n}{2}(0 + n - 1) = \frac{n(n-1)}{2}$

4. Combine the sums and the initial value to find an explicit formula for $u_n$.

Substitute the evaluated sums back into the overall equation, along with the given initial value $u_0 = 1$:
$u_n - 1 = (2^{n+1} - 2) + \frac{n(n-1)}{2}$
$u_n = 2^{n+1} - 1 + \frac{n(n-1)}{2}$

1. Identify the weekly testing sequence as an arithmetic progression and state its parameters.

In week 1, tests conducted = $300 \times 7 = 2100$.
In week 2, tests conducted = $400 \times 7 = 2800$.
The number of weekly tests forms an arithmetic progression with first term $a = 2100$ and common difference $d = 2800 - 2100 = 700$.

2. Formulate an inequality for the sum of tests over $n$ weeks.

Let $S_n$ be the total number of tests conducted in $n$ weeks.
We require $S_n \ge 200\,000$.
Using the sum formula $S_n = \frac{n}{2}[2a + (n-1)d]$:
$\frac{n}{2}[2(2100) + (n-1)(700)] \ge 200\,000$
$\frac{n}{2}[4200 + 700n - 700] \ge 200\,000$
$\frac{n}{2}[700n + 3500] \ge 200\,000$
$350n^2 + 1750n - 200\,000 \ge 0$

3. Solve the quadratic inequality to find the least integer number of weeks.

Dividing by 350:
$n^2 + 5n - \frac{4000}{7} \ge 0$
$7n^2 + 35n - 4000 \ge 0$

Solving $7n^2 + 35n - 4000 = 0$ using the quadratic formula:
$n = \frac{-35 \pm \sqrt{35^2 - 4(7)(-4000)}}{2(7)}$
$n = \frac{-35 \pm \sqrt{113225}}{14}$
Taking the positive root:
$n \approx \frac{-35 + 336.489}{14} \approx 21.53$

Since $n$ must be an integer, the least number of weeks is $22$.

1. Model the total percentage of the population vaccinated as a geometric series and find the sum to infinity.

Let the total population be $P$. The number of people inoculated each week forms a geometric progression with first term $a = 0.1P$ and common ratio $r = 0.9$.
The theoretical maximum number of people that can be inoculated is given by the sum to infinity of this series:
$S_\infty = \frac{a}{1 - r} = \frac{0.1P}{1 - 0.9} = \frac{0.1P}{0.1} = P$
This confirms the series converges to exactly 100% of the population.

2. Evaluate the feasibility by considering the time required to reach this theoretical limit.

The total number of people vaccinated after $n$ weeks is given by $S_n = \frac{0.1P(1 - 0.9^n)}{1 - 0.9} = P(1 - 0.9^n)$.
For the whole population to be fully inoculated, we require $P(1 - 0.9^n) = P$, which simplifies to $0.9^n = 0$.
This equation has no finite solution for $n$ and is only satisfied as $n \to \infty$. Therefore, it would take an infinite amount of time (weeks) to inoculate the whole population, making the proposal practically unfeasible.