| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2018 |
| Session | March |
| Marks | 5 |
| Topic | Sequences and series, recurrence and convergence |
| Type | Simple recurrence evaluation |
| Difficulty | Standard +0.3 This is a straightforward multi-part question on geometric sequences and limits. Part (i) requires recognizing a simple geometric sequence with first term √3 and common ratio (√3+1). Part (ii) involves explaining why (√3-1)^n → 0 since |√3-1| < 1. Part (iii) combines the results using algebraic manipulation. All parts are routine applications of standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 4.06b Method of differences: telescoping series8.01a 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 |
|---|---|---|
| M1 | 2.1 | Constructing or recognising a GP sequence |
| A1 | 1.1 | General term |
| [2] |
| Answer | Marks |
|---|---|
| B1 | 2.4 |
| [1] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | 2.4 | |
| A1 | 2.1 | oe |
| [2] |
## Part (i)
**Answer:** $A_2 = \sqrt{3}(\sqrt{3}+1), A_3 = \sqrt{3}(\sqrt{3}+1), \ldots$
$A_n = \sqrt{3}(\sqrt{3}+1)^{-1}$
| **M1** | **2.1** | Constructing or recognising a GP sequence |
| **A1** | **1.1** | General term |
| **[2]** | | |
## Part (ii)
**Answer:** $(0 < )\sqrt{3} - 1 = 0.732\ldots < 1 \Rightarrow (\sqrt{3}-1)^n \to 0$ as $n \to \infty$
| **B1** | **2.4** | |
| **[1]** | | |
## Part (iii)
**Answer:** $C_n \to \frac{\sqrt{3}(\sqrt{3}+1)^{n-1}}{\frac{1}{\sqrt{3}}(\sqrt{3}+1)^n} = \frac{3}{\sqrt{3}+1}$
| **M1** | **2.4** | |
| **A1** | **2.1** | oe |
| **[2]** | | |
---3 In this question you must show detailed reasoning.
(i) The sequence $\left\{ A _ { n } \right\}$ is given by $A _ { 1 } = \sqrt { 3 }$ and $A _ { n + 1 } = ( \sqrt { 3 } + 1 ) A _ { n }$ for $n \geqslant 1$.
Find an expression for $A _ { n }$ in terms of $n$.\\
(ii) The sequence $\left\{ B _ { n } \right\}$ is given by the formula
$$B _ { n } = \frac { 1 } { \sqrt { 3 } } \left( ( \sqrt { 3 } + 1 ) ^ { n } - ( \sqrt { 3 } - 1 ) ^ { n } \right) \text { for } n \geqslant 1 .$$
Explain why $B _ { n } \rightarrow \frac { 1 } { \sqrt { 3 } } ( \sqrt { 3 } + 1 ) ^ { n }$ as $n \rightarrow \infty$.\\
(iii) The sequence $\left\{ C _ { n } \right\}$ converges and is defined by $C _ { n } = \frac { A _ { n } } { B _ { n } }$ for $n \geqslant 1$.
Identify the limit of $C n$ as $n \rightarrow \infty$.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2018 Q3 [5]}}