| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2024 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Proof by induction |
| Type | Prove sequence property via recurrence |
| Difficulty | Standard +0.8 This is a structured induction proof requiring manipulation of the Fibonacci recurrence relation to establish a pattern (part a), then applying induction with a non-trivial inductive step. While the framework is guided, students must handle the recurrence algebra correctly and recognize how to use F_{n+4} = 3F_n + 2F_{n-1} to show divisibility by 3. This is moderately challenging—harder than routine induction proofs but accessible to well-prepared Further Maths students. |
| Spec | 4.01a Mathematical induction: construct proofs8.01e Fibonacci: and related sequences (e.g. Lucas numbers) |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (a) | b and d are the direction vectors of l … |
| Answer | Marks |
|---|---|
| a – c = µb or a – c = γd or (a – c) b = 0 or (a – c) d = 0 | B1 |
| Answer | Marks |
|---|---|
| [4] | 1.2 |
| Answer | Marks |
|---|---|
| 2.4 | or statement that one of b, d is a multiple of the other |
| Answer | Marks | Guidance |
|---|---|---|
| 4 | (b) | (a – c) d = 0 since d, a – c parallel (from (a)) |
| Answer | Marks |
|---|---|
| since a d is perpendicular to a | M1 |
| Answer | Marks |
|---|---|
| A1 | 1.1 |
| Answer | Marks |
|---|---|
| 2.4 | Correct answer, fully justified |
| Answer | Marks |
|---|---|
| (a d) + ( b d) – (c d) = 0 | M1 |
| But b d = 0 since b, d parallel a d = c d | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| since a d is perpendicular to a | A1 | Correct answer, fully justified |
Question 4:
4 | (a) | b and d are the direction vectors of l …
… so b = md for some scalar m or b d=0
a and c are (the position vectors of) points on l …
a – c = µb or a – c = γd or (a – c) b = 0 or (a – c) d = 0 | B1
B1
B1
B1
[4] | 1.2
2.2a
1.2
2.4 | or statement that one of b, d is a multiple of the other
or that they are parallel
or (a – c) is a multiple of (or parallel to) b (or d)
SC1 (a + b – c ) d = 0
4 | (b) | (a – c) d = 0 since d, a – c parallel (from (a))
a d = c d
a . (c d) = a . (a d) = 0
since a d is perpendicular to a | M1
A1
A1 | 1.1
1.1
2.4 | Correct answer, fully justified
ALT. Substitution for r and use of the Distributive property
of the VP:
(a d) + ( b d) – (c d) = 0 | M1
But b d = 0 since b, d parallel a d = c d | A1
a . (c d) = a . (a d) = 0
since a d is perpendicular to a | A1 | Correct answer, fully justified
[3]
ALT. Substitution for r and use of the Distributive property
of the VP:4 The first five terms of the Fibonacci sequence, $\left\{ \mathrm { F } _ { \mathrm { n } } \right\}$, where $n \geqslant 1$, are $F _ { 1 } = 1 , F _ { 2 } = 1 , F _ { 3 } = 2 , F _ { 4 } = 3$ and $F _ { 5 } = 5$.
\begin{enumerate}[label=(\alph*)]
\item Use the recurrence definition of the Fibonacci sequence, $\mathrm { F } _ { \mathrm { n } + 1 } = \mathrm { F } _ { \mathrm { n } } + \mathrm { F } _ { \mathrm { n } - 1 }$, to express $\mathrm { F } _ { \mathrm { n } + 4 }$ in terms of $\mathrm { F } _ { \mathrm { n } }$ and $\mathrm { F } _ { \mathrm { n } - 1 }$.
\item Hence prove by induction that $\mathrm { F } _ { \mathrm { n } }$ is a multiple of 3 when $n$ is a multiple of 4 .
\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2024 Q4 [5]}}