一种名为“Painomore”的药物在血液中的半衰期为 4 小时，其中药物的半衰期定义为药物在血液中减少 50% 所需的时间。该药物的配方规定为每 12 小时给药一次，剂量为 $D$ 毫克。

1. Determine the number of half-lives that elapse between the first and second dose

Time between doses $= 12$ hours
Number of half-lives $= \frac{12}{4} = 3$

2. Calculate the amount of drug left just before the second dose

Amount of drug left $= D \times \left(\frac{1}{2}\right)^3$
$= \frac{1}{8}D$

1. Determine the amount of drug after successive doses by identifying the common ratio.

From the half-life of 4 hours, the proportion of drug remaining after 12 hours is $\left(\frac{1}{2}\right)^{\frac{12}{4}} = \left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
Amount after 1st dose, $U_1 = D$
Amount after 2nd dose, $U_2 = D + \frac{1}{8}U_1 = D + \frac{1}{8}D$
Amount after 3rd dose, $U_3 = D + \frac{1}{8}U_2 = D + \frac{1}{8}D + \left(\frac{1}{8}\right)^2 D$
By extension, after the $n^{th}$ dose:
$U_n = D + \frac{1}{8}D + \left(\frac{1}{8}\right)^2 D + \dots + \left(\frac{1}{8}\right)^{n-1} D$

2. Apply the sum formula for a finite geometric series.

This is a geometric series with first term $a = D$, common ratio $r = \frac{1}{8}$, and $n$ terms.
Using the sum of a geometric progression $S_n = \frac{a(1-r^n)}{1-r}$:
$U_n = \frac{D\left(1 - \left(\frac{1}{8}\right)^n\right)}{1 - \frac{1}{8}}$

3. Simplify to match the given form and determine the constant $k$.

$U_n = \frac{D\left(1 - \left(\frac{1}{8}\right)^n\right)}{\frac{7}{8}}$
$U_n = \frac{8}{7}D\left(1 - \left(\frac{1}{8}\right)^n\right)$
Comparing this with $U_n = kD\left(1 - \left(\frac{1}{8}\right)^n\right)$, we have $k = \frac{8}{7}$.

1. Find the exact constant $k$ by identifying $U_n$ as the sum of a geometric series (required from part ii).

From the context, the half-life is 4 hours, so in 12 hours the drug reduces by a factor of $\left(\frac{1}{2}\right)^3 = \frac{1}{8}$.
The amount in the bloodstream immediately after the $n$-th dose forms a geometric series with first term $a = D$ and common ratio $r = \frac{1}{8}$.
$U_n = D + \frac{1}{8}D + \dots + \left(\frac{1}{8}\right)^{n-1}D$
Using the formula for the sum of a finite geometric series:
$U_n = \frac{D\left(1 - \left(\frac{1}{8}\right)^n\right)}{1 - \frac{1}{8}}$
$U_n = \frac{8}{7}D\left(1 - \left(\frac{1}{8}\right)^n\right)$
Thus, the constant $k = \frac{8}{7}$.

2. Determine the long-term maximum amount of drug in the bloodstream and solve for $D$.

Over a long period of time as the doses are given repeatedly, $n \to \infty$, which implies $\left(\frac{1}{8}\right)^n \to 0$.
The maximum amount of drug in the bloodstream (occurring immediately after each dose) converges to the sum to infinity:
$U_\infty = \frac{8}{7}D$
To prevent the drug level from ever being toxic, this long-term maximum must not exceed 60 milligrams.
$\frac{8}{7}D \le 60$
$D \le 60 \times \frac{7}{8}$
$D \le 52.5$

1. Substitute $D=40$ and the constant $k=\frac{8}{7}$ found in part (ii) into the formula for $U_n$.

$U_n = \frac{8}{7}(40)\left(1 - \left(\frac{1}{8}\right)^n\right) = \frac{320}{7}\left(1 - 8^{-n}\right)$

2. Formulate an inequality for $U_n$ to be within 4.3 mg of 50 mg.

We require $|U_n - 50| \le 4.3$, which translates to $45.7 \le U_n \le 54.3$.
Since the limiting maximum amount over time is $\frac{320}{7} \approx 45.71$, $U_n$ will always be strictly less than $54.3$.
Thus, the condition simplifies solely to $U_n \ge 45.7$:
$\frac{320}{7}\left(1 - 8^{-n}\right) \ge 45.7$

3. Solve the exponential inequality for $n$.

$1 - 8^{-n} \ge \frac{45.7 \times 7}{320}$
$1 - 8^{-n} \ge \frac{319.9}{320}$
$8^{-n} \le 1 - \frac{319.9}{320} = \frac{0.1}{320} = \frac{1}{3200}$
$8^n \ge 3200$
Taking $\ln$ on both sides:
$n \ln 8 \ge \ln 3200$
$n \ge \frac{\ln 3200}{\ln 8} \approx 3.88$

4. State the least integer value of doses.

Since $n$ represents the number of doses, it must be an integer.
The least integer $n$ satisfying $n \ge 3.88$ is $4$.

1. Form an equation to find the day when the dosage reaches the maximum of 20 mg.

The daily dosages form an arithmetic progression with first term $a = 4$ and common difference $d = 2$.
Let $n$ be the day the dosage reaches $20\text{ mg}$.
$u_n = a + (n - 1)d$
$4 + 2(n - 1) = 20$

2. Solve for $n$.

$2(n - 1) = 16$
$n - 1 = 8 \implies n = 9$
The dosage reaches $20\text{ mg}$ on the 9th day.

3. Calculate the total drug taken during the first 9 days.

Using the sum formula $S_n = \frac{n}{2}(a + l)$:
$S_9 = \frac{9}{2}(4 + 20)$
$S_9 = \frac{9}{2}(24) = 108\text{ mg}$

4. Calculate the drug taken over the remaining days and the final total amount.

The remaining days in the 14-day course $= 14 - 9 = 5\text{ days}$.
Drug taken during the remaining days $= 5 \times 20 = 100\text{ mg}$.
Total drug taken $= 108 + 100 = 208\text{ mg}$.