| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and series, recurrence and convergence |
| Type | Applied recurrence modeling |
| Difficulty | Standard +0.3 This is a straightforward recurrence relation question with standard parts: explaining a given model, proving a formula by induction (with the formula provided), and interpreting long-term behavior using the geometric term (1.1)^n. The induction is mechanical, and the convergence analysis only requires recognizing that 1.1^n → ∞. Slightly easier than average due to the scaffolded structure and context making the mathematics transparent. |
| Spec | 1.04e Sequences: nth term and recurrence relations4.01a Mathematical induction: construct proofs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(P_{n-1}\) is the population at the end of year \(n-1\) and this is increased by 10% by the end of year \(n\), so is multiplied by \(1.1\) to give \(1.1\times P_{n-1}\) as new population by natural causes | B1 | Need to see 10% increase linked to multiplication by scale factor 1.1 |
| \(Q\) is subtracted from \(1.1\times P_{n-1}\) as \(Q\) is the number of deer removed from the estate | B1 | Needs to explain that subtraction of \(Q\) indicates the removal of \(Q\) deer from population |
| So \(P_n=1.1P_{n-1}-Q\), \(P_0=5000\) as population at start is 5000 and \(n\in\mathbb{Z}^+\) | B1 | Needs complete explanation with mention of \(P_n=1.1P_{n-1}-Q\), \(P_0=5000\) being the initial number of deer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Let \(n=0\), then \(P_0=(5000-10Q)(1.1)^0+10Q=5000\), so result is true when \(n=0\) | B1 | Begins proof by induction by considering \(n=0\) |
| Assume result is true for \(n=k\): \(P_k=(1.1)^k(5000-10Q)+10Q\), then as \(P_{k+1}=1.1P_k-Q\), so \(P_{k+1}=\ldots\) | M1 | Assumes result is true for \(n=k\) and uses iterative formula to consider \(n=k+1\) |
| \(P_{k+1}=1.1\times1.1^k(5000-10Q)+1.1\times10Q-Q\) | A1 | Correct algebraic statement |
| So \(P_{k+1}=(5000-10Q)(1.1)^{k+1}+10Q\) | A1 | Correct statement for \(k+1\) in required form |
| Implies result holds for \(n=k+1\) and so by induction \(P_n=(5000-10Q)(1.1)^n+10Q\) is true for all integer \(n\) | B1 | Completes the inductive argument |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For \(Q<500\) the population of deer will grow, for \(Q>500\) the population of deer will fall | B1 | Consideration of both possible ranges of values for \(Q\) |
| For \(Q=500\) the population of deer remains steady at 5000 | B1 | Gives the condition for the steady state |
## Question 5:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $P_{n-1}$ is the population at the end of year $n-1$ and this is increased by 10% by the end of year $n$, so is multiplied by $1.1$ to give $1.1\times P_{n-1}$ as new population by natural causes | B1 | Need to see 10% increase linked to multiplication by scale factor 1.1 |
| $Q$ is subtracted from $1.1\times P_{n-1}$ as $Q$ is the number of deer removed from the estate | B1 | Needs to explain that subtraction of $Q$ indicates the removal of $Q$ deer from population |
| So $P_n=1.1P_{n-1}-Q$, $P_0=5000$ as population at start is 5000 and $n\in\mathbb{Z}^+$ | B1 | Needs complete explanation with mention of $P_n=1.1P_{n-1}-Q$, $P_0=5000$ being the initial number of deer |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $n=0$, then $P_0=(5000-10Q)(1.1)^0+10Q=5000$, so result is true when $n=0$ | B1 | Begins proof by induction by considering $n=0$ |
| Assume result is true for $n=k$: $P_k=(1.1)^k(5000-10Q)+10Q$, then as $P_{k+1}=1.1P_k-Q$, so $P_{k+1}=\ldots$ | M1 | Assumes result is true for $n=k$ and uses iterative formula to consider $n=k+1$ |
| $P_{k+1}=1.1\times1.1^k(5000-10Q)+1.1\times10Q-Q$ | A1 | Correct algebraic statement |
| So $P_{k+1}=(5000-10Q)(1.1)^{k+1}+10Q$ | A1 | Correct statement for $k+1$ in required form |
| Implies result holds for $n=k+1$ and so by induction $P_n=(5000-10Q)(1.1)^n+10Q$ is true for all integer $n$ | B1 | Completes the inductive argument |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For $Q<500$ the population of deer will grow, for $Q>500$ the population of deer will fall | B1 | Consideration of both possible ranges of values for $Q$ |
| For $Q=500$ the population of deer remains steady at 5000 | B1 | Gives the condition for the steady state |\begin{enumerate}
\item A population of deer on a large estate is assumed to increase by $10 \%$ during each year due to natural causes.
\end{enumerate}
The population is controlled by removing a constant number, $Q$, of the deer from the estate at the end of each year.
At the start of the first year there are 5000 deer on the estate.\\
Let $P _ { n }$ be the population of deer at the end of year $n$.\\
(a) Explain, in the context of the problem, the reason that the deer population is modelled by the recurrence relation
$$P _ { n } = 1.1 P _ { n - 1 } - Q , \quad P _ { 0 } = 5000 , \quad n \in \mathbb { Z } ^ { + }$$
(b) Prove by induction that $P _ { n } = ( 1.1 ) ^ { n } ( 5000 - 10 Q ) + 10 Q , \quad n \geqslant 0$\\
(c) Explain how the long term behaviour of this population varies for different values of $Q$.
\hfill \mbox{\textit{Edexcel FP2 AS Q5 [10]}}