| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure AS (Further Additional Pure AS) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Fibonacci and Related Sequences |
| Difficulty | Challenging +1.2 Part (a) requires straightforward recursive calculation of seven terms. Part (b)(i) involves pattern recognition by comparing sequences. Part (b)(ii) requires proof by induction, which is a standard Further Maths technique but applied to a non-homogeneous recurrence relation. While this is more sophisticated than basic A-level, it follows a predictable structure for students familiar with induction and Fibonacci sequences, making it moderately above average difficulty. |
| Spec | 4.01a Mathematical induction: construct proofs8.01a Recurrence relations: general sequences, closed form and recurrence8.01e Fibonacci: and related sequences (e.g. Lucas numbers) |
| Answer | Marks | Guidance |
|---|---|---|
| 3 | a | n 1 2 3 4 5 6 7 |
| Answer | Marks |
|---|---|
| n | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1 |
| 1.1 | 𝑢 = −1+0+𝑛−1 where 𝑛 = 1,2 |
| Answer | Marks |
|---|---|
| correct implies M1) | (Possibly BC) |
| Answer | Marks | Guidance |
|---|---|---|
| b | i | F 1 1 2 3 5 8 13 |
| Answer | Marks |
|---|---|
| n n | M1 |
| Answer | Marks |
|---|---|
| [2] | 1.1a |
| 2.2b | Any method for comparing the two |
| sequences | Listing elements of 𝑈 |
| Answer | Marks |
|---|---|
| n + 2 n + 1 n | B1 |
| Answer | Marks |
|---|---|
| A1 | 2.2a |
| Answer | Marks |
|---|---|
| 2.3 | Seen or used somewhere |
| Answer | Marks | Guidance |
|---|---|---|
| Given result correctly shown. | FT their result 3(b). | |
| Alternative method | B1 | Baseline case and assumption that |
| Answer | Marks | Guidance |
|---|---|---|
| 𝑘+1 𝑘+1 | M1 | Attempt at 𝑛 = 𝑘 +1 |
| Answer | Marks |
|---|---|
| 𝑘+2 | Inductive reasoning all correctly |
Question 3:
3 | a | n 1 2 3 4 5 6 7
U 0 −1 −1 −1 0 2 6
n | M1
A1
[2] | 1.1
1.1 | 𝑢 = −1+0+𝑛−1 where 𝑛 = 1,2
3
or 3 with visible substitution
All of 𝑈 to U correct (fully
3 7
correct implies M1) | (Possibly BC)
NB U and U are
1 2
given
b | i | F 1 1 2 3 5 8 13
n
U +n= F
n n | M1
A1
[2] | 1.1a
2.2b | Any method for comparing the two
sequences | Listing elements of 𝑈
𝑛
and 𝐹 is not sufficient
𝑛
F = U + n + 2 and F = U + n + 1
n + 2 n + 2 n + 1 n + 1
F = F + F
n + 2 n + 1 n
U = U + U + n − 1
n + 2 n + 1 n | B1
M1
A1 | 2.2a
2.1
2.3 | Seen or used somewhere
Their F , F and F substd.
n + 2 n + 1 n
into the usual Fibonacci r.r.
Given result correctly shown. | FT their result 3(b).
Alternative method | B1 | Baseline case and assumption that
statement is true for 𝑛 = 𝑘
𝑢 1 +1 = 𝐹 1 true and 𝑢 𝑘 +𝑘 = 𝐹 𝑘
Baseline case and assumption that
statement is true for 𝑛 = 𝑘
𝑈 +𝑘 +1 = 𝐹
𝑘+1 𝑘+1 | M1 | Attempt at 𝑛 = 𝑘 +1
𝐹 = 𝐹 + 𝐹 =
𝑘+2 𝑘+1 𝑘
𝑢 +𝑘 +1+𝑢 +𝑘 =
𝑘+1 𝑘
𝑢 −(𝑘 −1)+𝑘+1+𝑘
𝑘+2
= 𝑈 +𝑘 +2
𝑘+2 | Inductive reasoning all correctly
shown and conclusion
A1
[3]
𝐹 = 𝐹 + 𝐹 =
𝑘+2 𝑘+1 𝑘
𝑢 +𝑘 +1+𝑢 +𝑘 =
𝑘+1 𝑘
𝑢 −(𝑘 −1)+𝑘+1+𝑘
𝑘+2
= 𝑈 +𝑘 +2
𝑘+2
Inductive reasoning all correctly
shown and conclusion3 The sequence $\left\{ U _ { n } \right\}$ is given by $U _ { 1 } = 0 , U _ { 2 } = - 1$ and $U _ { n + 2 } = U _ { n + 1 } + U _ { n } + n - 1$ for $n \geqslant 1$.
\begin{enumerate}[label=(\alph*)]
\item List the first seven terms of this sequence.
The Fibonacci sequence $\left\{ \mathrm { F } _ { \mathrm { n } } \right\}$ is given by $\mathrm { F } _ { 1 } = 1 , \mathrm {~F} _ { 2 } = 1$ and $\mathrm { F } _ { \mathrm { n } + 2 } = \mathrm { F } _ { \mathrm { n } + 1 } + \mathrm { F } _ { \mathrm { n } }$ for $n \geqslant 1$.
\item \begin{enumerate}[label=(\roman*)]
\item By comparing the two sequences, give the relationship between $\mathrm { U } _ { \mathrm { n } }$ and $\mathrm { F } _ { \mathrm { n } }$.
\item Show that the relationship found in part (b)(i) holds for all $n \geqslant 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure AS 2022 Q3 [7]}}