| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2021 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Periodic Sequences |
| Difficulty | Challenging +1.8 This is a Further Maths question requiring analysis of recursive sequences with multiple cases. Students must set up and solve equations for constant sequences (straightforward), period-2 sequences (moderate algebraic manipulation), and period-4 sequences (requiring systematic checking that period-2 doesn't hold). The conceptual demand is significant but the algebraic techniques are standard, placing it well above average difficulty. |
| Spec | 8.01a Recurrence relations: general sequences, closed form and recurrence8.01c Sequence behaviour: periodic, convergent, divergent, oscillating, monotonic |
| Answer | Marks | Guidance |
|---|---|---|
| 9 | (a) | k |
| Answer | Marks |
|---|---|
| k = 16 | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| Answer | Marks |
|---|---|
| (b) | k 8 k |
| Answer | Marks |
|---|---|
| periodic with period 2 | M1 |
| Answer | Marks |
|---|---|
| [3] | 3.1a |
| Answer | Marks |
|---|---|
| 2.3 | u correct (simplified) and set = 2 |
| Answer | Marks |
|---|---|
| (c) | k k ( k + 4 8 ) k k (1 4 k + 2 8 8 ) |
| Answer | Marks |
|---|---|
| k = –18 | M1 |
| Answer | Marks |
|---|---|
| [5] | 1.1 |
| Answer | Marks |
|---|---|
| 2.2a | u and u attempted in terms of k |
Question 9:
9 | (a) | k
For constant sequence, set 2 = (from u = u )
6 + 2 2 1
k = 16 | M1
A1
[2] | 1.1a
1.1
(b) | k 8 k
u = = equated to u = 2
3 k 4 8 + k 1
6 +
8
Solving for k 96 + 2k = 8k k = 16
But this is the condition for u constant, so the sequence is never
n
periodic with period 2 | M1
A1
A1
[3] | 3.1a
1.1
2.3 | u correct (simplified) and set = 2
3
Correct conclusion stated with supporting reason
(c) | k k ( k + 4 8 ) k k (1 4 k + 2 8 8 )
u = = , u = =
4 8 k 2 8 8 + 1 4 k 5 k ( k + 4 8 ) k 2 + 1 3 2 k + 1 7 2 8
6 + 6 +
4 8 + k 2 8 8 + 1 4 k
For u periodic, period 4: u =u
n 5 1
14k2 + 288k = 2k2 + 264k +
3456
12(k2 + 2k – 288) = 0 (k – 16)(k + 18) = 0
k = –18 | M1
A1
M1
M1
A1
[5] | 1.1
1.1
3.1a
1.1
2.2a | u and u attempted in terms of k
4 5
At least u correct (simplified)
4
u must have been worked out and an algebraic
5
exp ression equated to 2
Solving a quadratic eqn. in k
Correct (single) answer only9 For each value of $k$ the sequence of real numbers $\left\{ u _ { n } \right\}$ is given by $u _ { 1 } = 2$ and $u _ { n + 1 } = \frac { k } { 6 + u _ { n } }$. For each of the following cases, either determine a value of $k$ or prove that one does not exist.
\begin{enumerate}[label=(\alph*)]
\item $\left\{ \mathrm { u } _ { n } \right\}$ is constant.
\item $\left\{ \mathrm { u } _ { \mathrm { n } } \right\}$ is periodic, with period 2 .
\item $\left\{ \mathrm { u } _ { \mathrm { n } } \right\}$ is periodic, with period 4 .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2021 Q9 [10]}}