| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2018 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Modelling with Recurrence Relations |
| Difficulty | Standard +0.3 This is a straightforward modelling question with recurrence relations. Part (a) requires explaining a given formula (minimal challenge), part (b) is a standard proof by induction with a simple recurrence relation, part (c) asks for a limit as nāā (routine), and part (d) involves solving a simple equation for equilibrium. All techniques are standard FP2 material with no novel insight required, making it slightly easier than average. |
| Spec | 4.01a Mathematical induction: construct proofs4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots8.01a Recurrence relations: general sequences, closed form and recurrence8.01b Induction: prove results for sequences and series8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(H_n\) is decreased by 3% so multiplied by \(0.97\) to give \(0.97H_n\) as new height after trimming | B1 | 3% decrease linked to scale factor of 0.97 |
| \(0.15\) is added to \(0.97H_n\) as \(0.15\) is 15cm in metres, representing yearly growth | B1 | Adding 0.15 must be referenced in metres |
| \(H_1 = 6\) is the height at start of year 1 before trimming | B1 | Explanation that \(H_1\) is the starting height of 6m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(n=1 \Rightarrow H_1 = (0.97)^{1-1}+5 = 6\), so true for \(n=1\) | B1 | Begins induction by considering \(n=1\) and obtains \(H_1=6\) |
| Assume true for \(n=k\) so \(H_k=(0.97)^{k-1}+5\), so \(H_{k+1}=0.97\big((0.97)^{k-1}+5\big)+0.15\) | M1 | Assumes true for \(n=k\) and uses iterative formula for \(n=k+1\) |
| \(H_{k+1}=(0.97)^k+4.85+0.15=(0.97)^k+5\) | A1 | Reaches \((0.97)^k+5\) with no errors |
| If true for \(n=k\) then true for \(n=k+1\), true for \(n=1\) so true for all positive integers \(n\) | B1 | Correct conclusion conveying all four underlined ideas |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The height will approach 5m | B1 | States height approaches 5m |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Require \(4 = 4x + 0.15\) | M1 | Correct strategy to find the required percentage |
| \(x = 0.9625\) so \(3.75\%\) | A1 | Interprets answer correctly in context |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $H_n$ is decreased by 3% so multiplied by $0.97$ to give $0.97H_n$ as new height after trimming | B1 | 3% decrease linked to scale factor of 0.97 |
| $0.15$ is added to $0.97H_n$ as $0.15$ is 15cm in metres, representing yearly growth | B1 | Adding 0.15 must be referenced in metres |
| $H_1 = 6$ is the height at start of year 1 before trimming | B1 | Explanation that $H_1$ is the starting height of 6m |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $n=1 \Rightarrow H_1 = (0.97)^{1-1}+5 = 6$, so true for $n=1$ | B1 | Begins induction by considering $n=1$ and obtains $H_1=6$ |
| Assume true for $n=k$ so $H_k=(0.97)^{k-1}+5$, so $H_{k+1}=0.97\big((0.97)^{k-1}+5\big)+0.15$ | M1 | Assumes true for $n=k$ and uses iterative formula for $n=k+1$ |
| $H_{k+1}=(0.97)^k+4.85+0.15=(0.97)^k+5$ | A1 | Reaches $(0.97)^k+5$ with no errors |
| If true for $n=k$ then true for $n=k+1$, true for $n=1$ so true for all positive integers $n$ | B1 | Correct conclusion conveying all four underlined ideas |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The height will approach 5m | B1 | States height approaches 5m |
## Part (d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Require $4 = 4x + 0.15$ | M1 | Correct strategy to find the required percentage |
| $x = 0.9625$ so $3.75\%$ | A1 | Interprets answer correctly in context |
---3 A tree at the bottom of a garden needs to be reduced in height. The tree is known to increase in height by 15 centimetres each year.
On the first day of every year, the height is measured and the tree is immediately trimmed by $3 \%$ of this height.
When the tree is measured, before trimming on the first day of year 1 , the height is 6 metres.\\
Let $H _ { n }$ be the height of the tree immediately before trimming on the first day of year $n$.
\begin{enumerate}[label=(\alph*)]
\item Explain, in the context of the problem, why the height of the tree may be modelled by the recurrence relation
$$H _ { n + 1 } = 0.97 H _ { n } + 0.15 , \quad H _ { 1 } = 6 , \quad n \in \mathbb { Z } ^ { + }$$
\item Prove by induction that $H _ { n } = 0.97 ^ { n - 1 } + 5 , \quad n \geqslant 1$
\item Explain what will happen to the height of the tree immediately before trimming in the long term.
\item By what fixed percentage should the tree be trimmed each year if the height of the tree immediately before trimming is to be 4 metres in the long term?
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 AS 2018 Q3 [10]}}