一种疾病的基本传染数 $R_0$ 是一个数学术语，用以表示一种传染病的传染性。它给出了一个患有该传染病的人平均会传染给其他人的数量。如果社区中没有人接种过该疾病的疫苗，或者没有人对该疾病具有免疫力，这种复制传播将会持续下去。 一种疾病的传染期是指感染者可以向其接触的任何易感者传播该疾病的时间段。例如，如果某种疾病的 $R_0$ 为 3，那么患有该疾病的人在传染期内平均会将该疾病传染给另外 3 个人。假设传染期为 2 周，且最初有 $x$ 人感染，则在第 2 周平均感染人数为 $3x$，在第 4 周为 $3^2x$，在第 6 周为 $3^3x$，依此类推。 在一个对该病毒没有免疫力的大量人群中，最初有 1 人感染了该病毒。已知该病毒的 $R_0$ 为 2.5，传染期为 2 周。

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.