| Exam Board | OCR |
|---|---|
| Module | Further Additional Pure (Further Additional Pure) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Sequences and Series |
| Type | Fibonacci and Related Sequences |
| Difficulty | Challenging +1.8 This is a Further Maths question on Binet's formula requiring multiple proof steps and conceptual understanding of sequences. Part (a) is routine Vieta's formulas, but parts (b)(ii)-(iii) require proving a recurrence relation and deducing integrality through induction reasoning, while part (c) demands insight into approximation and error analysis with the golden ratio. The multi-layered structure and need for mathematical maturity place it well above average difficulty. |
| Spec | 1.04e Sequences: nth term and recurrence relations8.01b Induction: prove results for sequences and series8.01e Fibonacci: and related sequences (e.g. Lucas numbers) |
| Answer | Marks | Guidance |
|---|---|---|
| 7 | (a) | + = 1 and = −1 |
| [1] | 1.1 | |
| (b) | i | S =2 +2 =(+)2 −2=3 |
| Answer | Marks |
|---|---|
| = 4 | B1 |
| Answer | Marks |
|---|---|
| [3] | 1.1 |
| Answer | Marks |
|---|---|
| 1.1 | Note that S = 1 from (a) (i) |
| Answer | Marks |
|---|---|
| ii | 𝑆 = 𝛼𝑛 + 2 +𝛽𝑛 + 2 |
| Answer | Marks |
|---|---|
| =(1)S n – (−1)S n – 1 = S n + S n – 1 | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.1a |
| 2.2a | Or using 𝛼2 −𝛼−1 = 0, 𝛽2 −𝛽 −1 = 0 |
| Answer | Marks |
|---|---|
| iii | Since S and S are integers, the Rec. Reln. of (a) (iii) |
| Answer | Marks | Guidance |
|---|---|---|
| induction) | B1 | |
| [1] | 2.4 | Explanation of the inductive nature of this scenario |
| (c) | i | It fails to give an integer |
| [1] | 3.5b | |
| ii | I N T ( n + 1 ) n e v e n |
| Answer | Marks |
|---|---|
| or as a carefully-defined “nearest integer” function | M1 |
| Answer | Marks |
|---|---|
| [2] | 3.5c |
| 3.5c | Use of the INT function |
Question 7:
7 | (a) | + = 1 and = −1 | B1
[1] | 1.1
(b) | i | S =2 +2 =(+)2 −2=3
2
( )
S =3 +3 =(+)2 +2 −(+)
3
= 4 | B1
M1
A1
[3] | 1.1
3.1a
1.1 | Note that S = 1 from (a) (i)
1
NB This is S =S S +S since = − 1
3 1 2 1
Using the exact values of 𝛼,𝛽 to find 𝑆 is accepted.
𝑖
ii | 𝑆 = 𝛼𝑛 + 2 +𝛽𝑛 + 2
𝑛 + 2
= (𝛼+𝛽)(𝛼𝑛+1 +𝛽𝑛+1)−𝛼𝑛+1𝛽
−𝛼𝛽𝑛+1
= (𝛼 +𝛽)(𝛼𝑛+1 +𝛽𝑛+1)−𝛼𝛽(𝛼𝑛 +𝛽𝑛)
=(1)S n – (−1)S n – 1 = S n + S n – 1 | M1
A1
[2] | 3.1a
2.2a | Or using 𝛼2 −𝛼−1 = 0, 𝛽2 −𝛽 −1 = 0
AG
iii | Since S and S are integers, the Rec. Reln. of (a) (iii)
1 2
S n an integer for all positive integers n (by
induction) | B1
[1] | 2.4 | Explanation of the inductive nature of this scenario
(c) | i | It fails to give an integer | B1
[1] | 3.5b
ii | I N T ( n + 1 ) n e v e n
S =
n I N T ( n ) n o d d
or as a carefully-defined “nearest integer” function | M1
A1
[2] | 3.5c
3.5c | Use of the INT function
Both correct. Note: INT(x + 1) = INT(x) + 17 Binet's formula for the $n$th Fibonacci number is given by $\mathrm { F } _ { \mathrm { n } } = \frac { 1 } { \sqrt { 5 } } \left( \alpha ^ { \mathrm { n } } - \beta ^ { \mathrm { n } } \right)$ for $n \geqslant 0$, where $\alpha$ and $\beta$ (with $\alpha > 0 > \beta$ ) are the roots of $x ^ { 2 } - x - 1 = 0$.
\begin{enumerate}[label=(\alph*)]
\item Write down the values of $\alpha + \beta$ and $\alpha \beta$.
\item Consider the sequence $\left\{ \mathrm { S } _ { \mathrm { n } } \right\}$, where $\mathrm { S } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } } + \beta ^ { \mathrm { n } }$ for $n \geqslant 0$.
\begin{enumerate}[label=(\roman*)]
\item Determine the values of $S _ { 2 }$ and $S _ { 3 }$.
\item Show that $S _ { n + 2 } = S _ { n + 1 } + S _ { n }$ for $n \geqslant 0$.
\item Deduce that $S _ { n }$ is an integer for all $n \geqslant 0$.
\end{enumerate}\item A student models the terms of the sequence $\left\{ \mathrm { S } _ { \mathrm { n } } \right\}$ using the formula $\mathrm { T } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } }$.
\begin{enumerate}[label=(\roman*)]
\item Explain why this formula is unsuitable for every $n \geqslant 1$.
\item Considering the cases $n$ even and $n$ odd separately, state a modification of the formula $\mathrm { T } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } }$, other than $\mathrm { T } _ { \mathrm { n } } = \alpha ^ { \mathrm { n } } + \beta ^ { \mathrm { n } }$, such that $\mathrm { T } _ { \mathrm { n } } = \mathrm { S } _ { \mathrm { n } }$ for all $n \geqslant 1$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Additional Pure 2023 Q7 [10]}}