| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 12 |
| Topic | Sequences and Series |
| Type | First-Order Linear Recurrence Relations |
| Difficulty | Challenging +1.2 This question on continued fractions requires understanding convergence to fixed points and solving quadratic equations, but follows a heavily scaffolded structure. Part (i) guides students through the method before applying it independently in parts (ii) and (iii). The algebraic manipulation is straightforward, and finding suitable a,b values in (iii) involves simple pattern recognition from the target form. While above average due to the unfamiliar context and multi-step reasoning, the scaffolding and routine algebraic techniques keep it accessible. |
| Spec | 8.01a Recurrence relations: general sequences, closed form and recurrence8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic8.01d Sequence limits: limit of nth term as n tends to infinity, steady-states |
| Answer | Marks | Guidance |
|---|---|---|
| B1 | 3.4 | BC from repeated use of \(u_{n+1} = 1 + \frac{1}{u_n}, u_0 = 1\) |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 1.1a | |
| A1 | 1.1 | Allow the positive root given only |
| E1 | 2.3 | Justification for taking the positive root |
| [3] |
| Answer | Marks |
|---|---|
| M1 | 1.1 |
| A1 | 1.1 |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 3.4 | No need to get it into standard quadratic form |
| M1 | 3.1a | Must be substituted fully |
| B1 | 1.1 | Noted at any stage |
| M1 | 2.1 | Comparing integer and surd components |
| M1, A1 | 3.1a, 1.1 | Solving simultaneously for \(a, b\) |
| [6] |
## Part (i)(a)
**Answer:** 1.618 034
| **B1** | **3.4** | BC from repeated use of $u_{n+1} = 1 + \frac{1}{u_n}, u_0 = 1$ |
| **[1]** | | |
## Part (i)(b)
**Answer:** $\omega = 1 + \frac{1}{\omega} \Rightarrow \omega^2 - \omega - 1 = 0$
$\Rightarrow \omega = \frac{1 \pm \sqrt{5}}{2}$
$\Rightarrow \omega = \frac{1+\sqrt{5}}{2}$ since all terms of the sequence are positive
| **M1** | **1.1a** | |
| **A1** | **1.1** | Allow the positive root given only |
| **E1** | **2.3** | Justification for taking the positive root |
| **[3]** | | |
## Part (ii)
**Answer:** For $\omega = \frac{1}{2+\omega}, \omega^2 + 2\omega - 1 = 0$
$\Rightarrow \omega = \sqrt{2} - 1$
| **M1** | **1.1** | |
| **A1** | **1.1** | |
| **[2]** | | |
## Part (iii)
**Answer:** For $\omega = a + \frac{1}{b+\omega}, \omega^2 + b\omega = ab + a\omega + 1$
Substituting $\omega = 2 + \sqrt{10}$
$\omega^2 = 14 + 4\sqrt{10}$
gives $14 + 4\sqrt{10} + 2b + b\sqrt{10} = ab + 2a + a\sqrt{10} + 1$
$\sqrt{10}: 4 = a - b$ and $\text{intr.}: 14 + 2b = ab + 2a + 1 \Rightarrow 5 = ab$
Solving simultaneously for $a, b$
$a = 5, b = 1$
| **M1** | **3.4** | No need to get it into standard quadratic form |
| **M1** | **3.1a** | Must be substituted fully |
| **B1** | **1.1** | Noted at any stage |
| **M1** | **2.1** | Comparing integer and surd components |
| **M1, A1** | **3.1a, 1.1** | Solving simultaneously for $a, b$ |
| **[6]** | | |7 Irrational numbers can be modelled by sequences $\left\{ u _ { n } \right\}$ of rational numbers of the form
$$u _ { 0 } = 1 \text { and } u _ { n + 1 } = a + \frac { 1 } { b + u _ { n } } \text { for } n \geqslant 0 \text {, }$$
where $a$ and $b$ are non-negative integer constants.
\begin{enumerate}[label=(\roman*)]
\item (a) The constants $a = 1$ and $b = 0$ produce the irrational number $\omega$. State the value of $\omega$ correct to six decimal places.\\
(b) By setting $u _ { n + 1 }$ and $u _ { n }$ equal to $\omega$, determine the exact value of $\omega$.
\item Use the method of part (i) (b) to find the exact value of the irrational number produced by taking $a = 0$ and $b = 2$.
\item Find positive integers $a$ and $b$ which would produce the irrational number $2 + \sqrt { 10 }$.
\section*{END OF QUESTION PAPER}
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q7 [12]}}