Arithmetic & Geometric Progressions: Disease spread: GP model & bed capacity — MI 2025 H2 Math Prelim Paper 1
What this question tests
Question
Epidemiologists are modelling the spread of an infectious disease in a town. The following assumptions are made:
- At Week 0, there are 500 new infections.
- In each subsequent week, the number of new infections is 60% the number of new infections in the previous week, due to ongoing community transmission.
Let \(u_n\) denote the number of new infections in Week \(n\), for \(n \geq 0\), \(n \in \mathbb{Z}\).
- Write down the value \(u_1\) and verify that \(u_2 = 180\).
- If \(S_n\) denotes the total cumulative number of new infections after \(n\) weeks, show that \(S_n = 1250\!\left(1-0.6^{n+1}\right)\).
The local healthcare facility has a capacity of 1240 beds. You may assume that each infected person occupies one bed, and that none are discharged during the period being considered.
- Find the smallest value of \(n\) such that the total number of infections up to and including Week \(n\) first exceeds 1240.
The healthcare facility later increases its capacity to 1280 beds.
- Based on the given model, comment on whether this new capacity is sufficient to accommodate all infected persons over a prolonged period of time.
Show full worked solution▾
(i) GP: \(a = 500\), \(r = 0.6\).
\(u_0 = 500\).
\(u_1 = 0.6\times 500 = 300\).
\(u_2 = 0.6\times(0.6\times 500) = 500\times 0.6^2 = 180\) (verified).
\(u_1 = 300\)(ii)
| Week \(n\) | New case, \(u_n\) | Total Cumulative, \(S_n\) |
|---|---|---|
| \(0\) | \(500\) | \(500\) |
| \(1\) | \(500 \times 0.6^1\) | \(500 + 500 \times 0.6^1\) |
| \(2\) | \(500 \times 0.6^2\) | \(500 + 500 \times 0.6^1 + 500 \times 0.6^2\) |
| \(\ldots\) | \(\ldots\) | \(\ldots\) |
| \(n\) | \(500 \times 0.6^n\) | \(500 + 500 \times 0.6^1 + 500 \times 0.6^2\) |
| \(\quad + \ldots + 500 \times 0.6^n\) |
(iii) \(1250(1-0.6^{n+1}) > 1240\).
| \(n\) | \(1250(1-0.6^{n+1})\) |
|---|---|
| 8 | 1237.4 |
| 9 | 1242.4 |
\(n = 9\).
(iv) \(S_n = 1250(1-0.6^{n+1})\).
As \(n\to\infty\), \(0.6^{n+1}\to 0\), so \(S_n\to 1250\).
Since the maximum cumulative infection is \(1250 < 1280\), the new patient capacity will be sufficient.