1. **Problem 1: Antibiotic Decay and Dosage** A doctor prescribes a 300-mg antibiotic tablet every 8 hours. After each dose, only 40% of the drug remains just before the next dose. 2. **Part A: Amount after 2nd and 3rd tablets** - After 1st tablet: $Q_1 = 300$ mg - Before 2nd tablet, drug decays to $0.4 \times 300 = 120$ mg - After 2nd tablet: $Q_2 = 120 + 300 = 420$ mg - Before 3rd tablet, drug decays to $0.4 \times 420 = 168$ mg - After 3rd tablet: $Q_3 = 168 + 300 = 468$ mg 3. **Part B: Recursion formula** - The drug amount after the $(n+1)$-th tablet is the decayed amount after the $n$-th tablet plus 300 mg: $$Q_{n+1} = 0.4 Q_n + 300$$ - Initial condition: $Q_1 = 300$ 4. **Part C: Closed-form formula using geometric series** - The recursion is a non-homogeneous linear recurrence with constant coefficients. - General solution: $$Q_n = 0.4^{n-1} Q_1 + 300 \sum_{k=0}^{n-2} 0.4^k$$ - Using geometric series sum formula: $$\sum_{k=0}^{m} r^k = \frac{1-r^{m+1}}{1-r}$$ - Substitute $r=0.4$, $Q_1=300$: $$Q_n = 300(0.4)^{n-1} + 300 \times \frac{1-(0.4)^{n-1}}{1-0.4} = 300(0.4)^{n-1} + 500(1-(0.4)^{n-1})$$ - Simplify: $$Q_n = 500 - 200(0.4)^{n-1}$$ - Alternatively: $$Q_n = 500(1 - 0.4^{n-1})$$ 5. **Part D: Long-term behavior** - As $n \to \infty$, since $0.4^n \to 0$: $$\lim_{n \to \infty} Q_n = 500(1 - 0) = 500$$ - So, the drug amount stabilizes at 500 mg immediately after a tablet is taken. --- 6. **Problem 2: Bird Population Model (Lack's Model)** - Number of birds at year $t$: $N(t)$ - Each bird lays $b$ eggs; proportion $s$ of offspring survive. - Adults die at end of year. 7. **Part A: Recursion for $N(t+1)$** - Number of surviving chicks: $$s \times b \times (b N(t)) = s b^2 N(t)$$ - Since adults die, only surviving chicks form next year's population: $$N(t+1) = s b^2 N(t)$$ 8. **Part B: Numerical values for $N(t)$ with $N(0)=10$, $b=5$, $s=0.1$** - Calculate $s b^2 = 0.1 \times 25 = 2.5$ - Recursion: $$N(t+1) = 2.5 N(t)$$ - Compute values: $t=0$: $10$ $t=1$: $10 \times 2.5 = 25$ $t=2$: $25 \times 2.5 = 62.5$ $t=3$: $62.5 \times 2.5 = 156.25$ $t=4$: $156.25 \times 2.5 = 390.625$ $t=5$: $390.625 \times 2.5 = 976.5625$ $t=6$: $976.5625 \times 2.5 = 2441.40625$ $t=7$: $2441.40625 \times 2.5 = 6103.515625$ $t=8$: $6103.515625 \times 2.5 = 15258.7890625$ $t=9$: $15258.7890625 \times 2.5 = 38146.97265625$ $t=10$: $38146.97265625 \times 2.5 = 95367.431640625$ 9. **Part D: Analytical expressions for $N(0)$ to $N(3)$** $$N(t) = 10 (2.5)^t$$ - $N(0) = 10 (2.5)^0 = 10$ - $N(1) = 10 (2.5)^1 = 25$ - $N(2) = 10 (2.5)^2 = 62.5$ - $N(3) = 10 (2.5)^3 = 156.25$ 10. **Part E: General solution and function type** - General solution: $$N(t) = N(0) (s b^2)^t$$ - This is an exponential function describing population growth. --- **Summary:** - Problem 1 models drug decay and dosage with a linear recurrence and geometric series. - Problem 2 models bird population growth with an exponential function.