| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2017 |
| Session | December |
| Marks | 10 |
| Topic | Sequences and Series |
| Type | Sequence Behaviour Classification |
| Difficulty | Challenging +1.2 This is a Further Maths question on recurrence relations requiring analysis of sequence behavior, finding fixed points, and proof by induction. While it involves multiple parts and some problem-solving (particularly the divisibility proof), the techniques are standard for Further Maths: part (i) is computational, part (ii) solves a quadratic equation for fixed points, and part (iii) is a straightforward induction proof. The question is moderately challenging but uses well-established methods without requiring novel insights. |
| Spec | 8.01a Recurrence relations: general sequences, closed form and recurrence8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(U_1 = 1, U_2 = 1^2 - 3 = -2, U_3 = (-2)^2 - 3 = 1, ...\) giving a periodic sequence with period 2 | B1 B1 [2] | 1.1 2.2a |
| (ii) Sequence constant if and only if \(a = a^2 - 6\) ⟹ \(a = 3\) or \(-2\) | M1 A1 [2] | 3.1a 1.1 |
| (iii) \(U_2 = (-1)^2 - 8 = -7\) is divisible by 7 (hence result is true for \(n = 2\)); Suppose that \(U_{2k} = 7m\) (i.e. the result is true for \(n = 2k\)); Then \(U_{2k+1} = (7m)^2 - 8 = 49m^2 - 8\); so \(U_{2(k+1)} = (49m^2 - 8)^2 - 8 = 2401m^4 - 784m^2 + 56 = 7(343m^4 - 112m^2 + 8)\); Explanation that result is true for \(n = 2\) and true for \(n = 2(k + 1)\) if true for \(n = 2k\) so that the result holds for all even terms of the sequence by induction | B1 M1 M1 M1 A1 E1 [6] | 1.1 2.1 2.1 3.1a 2.1 2.4 |
| Answer | Marks |
|---|---|
| \(U_2 = (-1)^2 - 8 = -7\) is divisible by 7 (hence result is true for \(n = 2\)); Writing \(U_{2k} = U, U_{2k+1} = U^2 - 8\) and \(U_{2k+2} = (U^2 - 8)^2 - 8 = U^4 - 16U^2 + 56\); Each term is now a multiple of 7 (the first two by hypothesis) hence ... explanation as above | B1 M1 M1A1 M1 E1 |
| Answer | Marks |
|---|---|
| Working modulo 7: \(U_1 \equiv -1 \Rightarrow U_2 \equiv (-1)^2 - 8 \equiv 0\); ⟹ \(U_3 \equiv 0^2 - 8 \equiv -1\); The sequence alternates −1, 0, −1, 0, ... (mod 7) and all the even-numbered terms are divisible by 7 | M1 M1A1 A1 E1 A1 |
**(i)** $U_1 = 1, U_2 = 1^2 - 3 = -2, U_3 = (-2)^2 - 3 = 1, ...$ giving a periodic sequence with period 2 | B1 B1 [2] | 1.1 2.2a | Accept clear statement that sequence alternates 1, −2, 1, −2, 1, ...
**(ii)** Sequence constant if and only if $a = a^2 - 6$ ⟹ $a = 3$ or $-2$ | M1 A1 [2] | 3.1a 1.1 | Use $U_n = U_{n+1} = a$; BC
**(iii)** $U_2 = (-1)^2 - 8 = -7$ is divisible by 7 (hence result is true for $n = 2$); Suppose that $U_{2k} = 7m$ (i.e. the result is true for $n = 2k$); Then $U_{2k+1} = (7m)^2 - 8 = 49m^2 - 8$; so $U_{2(k+1)} = (49m^2 - 8)^2 - 8 = 2401m^4 - 784m^2 + 56 = 7(343m^4 - 112m^2 + 8)$; Explanation that result is true for $n = 2$ and true for $n = 2(k + 1)$ if true for $n = 2k$ so that the result holds for all even terms of the sequence by induction | B1 M1 M1 M1 A1 E1 [6] | 1.1 2.1 2.1 3.1a 2.1 2.4 | No need to specify that $m \in \mathbb{Z}$; Calculating $U_{2k+1}$; Attempt at $U_{2(k+1)}$ in terms of $U_{2k}$; Collecting terms suitably to show that $U_{2k+2}$ is a multiple of 7
**or**
$U_2 = (-1)^2 - 8 = -7$ is divisible by 7 (hence result is true for $n = 2$); Writing $U_{2k} = U, U_{2k+1} = U^2 - 8$ and $U_{2k+2} = (U^2 - 8)^2 - 8 = U^4 - 16U^2 + 56$; Each term is now a multiple of 7 (the first two by hypothesis) hence ... explanation as above | B1 M1 M1A1 M1 E1
**or**
Working modulo 7: $U_1 \equiv -1 \Rightarrow U_2 \equiv (-1)^2 - 8 \equiv 0$; ⟹ $U_3 \equiv 0^2 - 8 \equiv -1$; The sequence alternates −1, 0, −1, 0, ... (mod 7) and all the even-numbered terms are divisible by 7 | M1 M1A1 A1 E1 A16 For real constants $a$ and $b$, the sequence $U _ { 1 } , U _ { 2 } , U _ { 3 } , \ldots$ is given by
$$U _ { 1 } = a \text { and } U _ { n } = \left( U _ { n - 1 } \right) ^ { 2 } - b \text { for } n \geqslant 2 .$$
(i) Determine the behaviour of the sequence in the case where $a = 1$ and $b = 3$.\\
(ii) In the case where $b = 6$, find the values of $a$ for which the sequence is constant.\\
(iii) In the case where $a = - 1$ and $b = 8$, prove that $U _ { n }$ is divisible by 7 for all even values of $n$.
\hfill \mbox{\textit{OCR Further Additional Pure AS 2017 Q6 [10]}}