C1 June 2018 Q6
6. A sequence \(a_1, a_2, a_3, \ldots\) is defined by\[\begin{aligned} a_1 &= 4 \\ a_{n+1} &= \frac{a_n}{a_n + 1}, \qquad n \geqslant 1,\ n \in \mathbb{N} \end{aligned}\]
Write your answers as simplified fractions. (3)
Given that\[a_n = \frac{4}{pn + q}, \text{ where } p \text{ and } q \text{ are constants}\]
| Scheme | Marks |
|---|---|
| \(a_1 = 4 \Rightarrow a_2 = \dfrac{4}{4 + 1}\) | M1 |
| \(\dfrac{4}{5}, \dfrac{4}{9}, \dfrac{4}{13}\) | A1A1 |
| (3) |
Notes
M1: Attempts to use the given recurrence relation correctly at least once e.g. \(a_2 = \dfrac{4}{4 + 1}\) or \(a_3 = \dfrac{\text{their } a_2}{\left(\text{their } a_2\right) + 1}\) or \(a_4 = \dfrac{\text{their } a_3}{\left(\text{their } a_3\right) + 1}\).
May be implied by their term(s).
A1: Two of \(\dfrac{4}{5}, \dfrac{4}{9}, \dfrac{4}{13}\) which may be un-simplified. Accept for example \(0.8, \dfrac{0.8}{1.8}, \ldots\) or \(\dfrac{4}{5}, \dfrac{\frac{4}{5}}{1 + \frac{4}{5}}, \ldots\)
A1: \(\dfrac{4}{5}, \dfrac{4}{9}, \dfrac{4}{13}\) (Allow 0.8 for \(\dfrac{4}{5}\))
| Scheme | Marks |
|---|---|
| \(p = 4\) or e.g. \(4 = \dfrac{4}{p + q},\ \text{"}\dfrac{4}{5}\text{"} = \dfrac{4}{2p + q}\) \(\Rightarrow p = \ldots\) or \(q = \ldots\) | M1 |
| \(a_n = \dfrac{4}{4n - 3} \Rightarrow p = 4\) and \(q = -3\) | A1 |
| Correct answer only scores both marks. | |
| (2) |
Notes
M1: \(a_n = \dfrac{4}{4n \pm \ldots}\) or \(p = 4\) OR
Uses 2 terms to set up and solve two correct equations for their fractions in \(p\) and \(q\) to obtain a value for \(p\) or a value for \(q\).
A1: Either \(a_n = \dfrac{4}{4n - 3}\) OR \(p = 4\) and \(q = -3\)
| Scheme | Marks |
|---|---|
| \(\dfrac{4}{4N - 3} = \dfrac{4}{321} \Rightarrow N = \ldots\) | M1 |
| \(\left(N =\right)81\) | A1 |
| Allow trial and improvement if 81 is clearly identified and then award both marks following a correct answer in (b) but just trying random values is M0 | |
| (2) | |
| (7 marks) |
Notes
M1: Solves their \(\dfrac{4}{pN + q} = \dfrac{4}{321}\) to obtain a value for \(N\) or \(n\).
A1: Cao (ignore what they use for \(N\))