| Exam Board | Edexcel |
|---|---|
| Module | FP2 AS (Further Pure 2 AS) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Modelling with Recurrence Relations |
| Difficulty | Standard +0.3 This is a straightforward application of first-order linear recurrence relations with a constant term. Part (a) requires explaining a given model (minimal challenge), part (b) asks for a simple assumption, part (c) involves standard technique (complementary function + particular solution), and part (d) is direct substitution. While it requires multiple steps, all techniques are routine for FP2 students with no novel problem-solving required. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots8.01a Recurrence relations: general sequences, closed form and recurrence8.01f First-order recurrence: solve using auxiliary equation and complementary function |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(U_{n-1}\) is the amount in the saving account \(n-1\) years after Jim's 11th birthday. This is increased by 2% each year, so is multiplied by 1.02 to give \(1.02U_{n-1}\) | B1 | 3.3 — Need to explain that 2% interest rate linked to multiplication by scale factor 1.02 |
| Jim's parents invest £500 for each subsequent birthday so 500 is added | B1 | 3.4 — Need to explain that 500 is added due to receiving £500 each year |
| \(U_0 = 1000\) as this is the amount invested on Jim's 11th birthday | B1 | 1.1b — Needs to explain that \(U_0 = 1000\) is the initial amount invested |
| (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| One of, for example: the interest rate stays the same each year; Jim does not withdraw any money from the savings account; Jim only saves the birthday money +£500 in this saving account, he does not invest any other money. | B1 | 3.5b |
| (1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| A complete method to solve the recurrence relation using \(U_n = \text{CF} + \text{PS} = c(1.02)^n + \lambda\) | M1 | 3.1a |
| \(\text{PS} = \lambda \Rightarrow \lambda = 1.02\lambda + 500\) leading to \(\lambda = \ldots\) | M1 | 1.1b — Uses PS \(= \lambda\) to find a value for \(\lambda\) |
| \(\lambda = -25\,000\) | A1 | 1.1b |
| Uses \(U_0 = 1000\) and their value for \(\lambda\) to find the value of \(c\): \(1000 = c(1.02)^0 - 25\,000\), giving \(c = \ldots(26\,000)\) | M1 | 1.1b |
| \(U_n = 26\,000(1.02)^n - 25\,000 \quad (n \geq 0)\) | A1 | 1.1b — Fully correctly defined sequence |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Realises that \(U_n =\) term of a GP \(+\) sum of a GP both with \(r = 1.02\) | M1 | 3.1a — A correct form for \(U_n\) |
| Sum of a GP \(= \dfrac{500(1-1.02^n)}{1-1.02}\) or \(\dfrac{500(1.02^n - 1)}{1.02 - 1}\) | M1, A1 | 1.1b, 1.1b — For sum of GP with \(a=500\), \(r=1.02\) and uses \(n\) or \(n-1\); correct sum |
| Term of a GP \(= 1000(1.02)^n\) or \(1000(1.02)^{n-1}\) | M1 | 1.1b — For term of GP with \(a=1000\), \(r=1.02\) |
| \(U_n = 1000(1.02)^n - 25\,000(1-1.02^n)\) or \(U_n = 1000(1.02)^n + 25\,000(1.02^n - 1)\) | A1 | 1.1b — Fully correctly defined sequence \(U_n\) |
| (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Uses \(U_n = 26\,000(1.02)^n - 25\,000\), with either \(n = 7\) or \(n = 8\) | M1 | 3.4 |
| \(U_7 = 4865.83 > 4500\), therefore Jim will have enough money in his savings account to buy a car costing £4500. | A1ft | 2.2a — Finds \(U_7\), compares with 4500 and comes to an appropriate conclusion. Follow through on their value of \(U_7\) |
| (2) | ||
| (11 marks total) |
## Question 5:
---
### Part 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $U_{n-1}$ is the amount in the saving account $n-1$ years after Jim's 11th birthday. This is increased by 2% each year, so is multiplied by 1.02 to give $1.02U_{n-1}$ | B1 | 3.3 — Need to explain that 2% interest rate linked to multiplication by scale factor 1.02 |
| Jim's parents invest £500 for each subsequent birthday so 500 is added | B1 | 3.4 — Need to explain that 500 is added due to receiving £500 each year |
| $U_0 = 1000$ as this is the amount invested on Jim's 11th birthday | B1 | 1.1b — Needs to explain that $U_0 = 1000$ is the initial amount invested |
| | **(3)** | |
---
### Part 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| One of, for example: the interest rate stays the same each year; Jim does not withdraw any money from the savings account; Jim only saves the birthday money +£500 in this saving account, he does not invest any other money. | B1 | 3.5b |
| | **(1)** | |
---
### Part 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| A complete method to solve the recurrence relation using $U_n = \text{CF} + \text{PS} = c(1.02)^n + \lambda$ | M1 | 3.1a |
| $\text{PS} = \lambda \Rightarrow \lambda = 1.02\lambda + 500$ leading to $\lambda = \ldots$ | M1 | 1.1b — Uses PS $= \lambda$ to find a value for $\lambda$ |
| $\lambda = -25\,000$ | A1 | 1.1b |
| Uses $U_0 = 1000$ and their value for $\lambda$ to find the value of $c$: $1000 = c(1.02)^0 - 25\,000$, giving $c = \ldots(26\,000)$ | M1 | 1.1b |
| $U_n = 26\,000(1.02)^n - 25\,000 \quad (n \geq 0)$ | A1 | 1.1b — Fully correctly defined sequence |
| | **(5)** | |
**Alternative 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Realises that $U_n =$ term of a GP $+$ sum of a GP both with $r = 1.02$ | M1 | 3.1a — A correct form for $U_n$ |
| Sum of a GP $= \dfrac{500(1-1.02^n)}{1-1.02}$ or $\dfrac{500(1.02^n - 1)}{1.02 - 1}$ | M1, A1 | 1.1b, 1.1b — For sum of GP with $a=500$, $r=1.02$ and uses $n$ or $n-1$; correct sum |
| Term of a GP $= 1000(1.02)^n$ or $1000(1.02)^{n-1}$ | M1 | 1.1b — For term of GP with $a=1000$, $r=1.02$ |
| $U_n = 1000(1.02)^n - 25\,000(1-1.02^n)$ or $U_n = 1000(1.02)^n + 25\,000(1.02^n - 1)$ | A1 | 1.1b — Fully correctly defined sequence $U_n$ |
| | **(5)** | |
---
### Part 5(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Uses $U_n = 26\,000(1.02)^n - 25\,000$, with either $n = 7$ or $n = 8$ | M1 | 3.4 |
| $U_7 = 4865.83 > 4500$, therefore Jim will have enough money in his savings account to buy a car costing £4500. | A1ft | 2.2a — Finds $U_7$, compares with 4500 and comes to an appropriate conclusion. Follow through on their value of $U_7$ |
| | **(2)** | |
| | **(11 marks total)** | |\begin{enumerate}
\item On Jim's 11 th birthday his parents invest $\pounds 1000$ for him in a savings account.
\end{enumerate}
The account earns 2\% interest each year.\\
On each subsequent birthday, Jim's parents add another $\pounds 500$ to this savings account.\\
Let $U _ { n }$ be the amount of money that Jim has in his savings account $n$ years after his 11th birthday, once the interest for the previous year has been paid and the $\pounds 500$ has been added.\\
(a) Explain, in the context of the problem, why the amount of money that Jim has in his savings account can be modelled by the recurrence relation of the form
$$U _ { n } = 1.02 U _ { n - 1 } + 500 \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$
(b) State an assumption that must be made for this model to be valid.\\
(c) Solve the recurrence relation
$$U _ { n } = 1.02 U _ { n - 1 } + 500 \quad U _ { 0 } = 1000 \quad n \in \mathbb { Z } ^ { + }$$
Jim hopes to be able to buy a car on his 18th birthday.\\
(d) Use the answer to part (c) to find out whether Jim will have enough money in his savings account to buy a car that costs $\pounds 4500$
\hfill \mbox{\textit{Edexcel FP2 AS 2019 Q5 [11]}}