OCR Further Additional Pure 2022 June — Question 3 6 marks

Exam BoardOCR
ModuleFurther Additional Pure (Further Additional Pure)
Year2022
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyChallenging +1.2 This is a structured induction proof with a given formula to verify. While it involves the golden ratio and Fibonacci numbers (conceptually interesting), the proof follows a standard template: verify base cases, assume for n and n-1, then show for n+1 using the recurrence relation and the property φ² = φ + 1. The algebraic manipulation is straightforward once the setup is recognized. Slightly above average due to the non-trivial formula and need to use φ's defining property, but well within reach for Further Maths students familiar with induction.
Spec4.01a Mathematical induction: construct proofs

3 The irrational number \(\phi = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )\) plays a significant role in the sequence of Fibonacci numbers given by \(\mathrm { F } _ { 0 } = 0 , \mathrm {~F} _ { 1 } = 1\) and \(\mathrm { F } _ { \mathrm { n } + 1 } = \mathrm { F } _ { \mathrm { n } } + \mathrm { F } _ { \mathrm { n } - 1 }\) for \(n \geqslant 1\). Prove by induction that, for each positive integer \(n , \phi ^ { n } = \mathrm { F } _ { \mathrm { n } } \times \phi + \mathrm { F } _ { \mathrm { n } - 1 }\).

Question 3:
AnswerMarks
31 = 1 . + 0 = ( F ) + ( F )    so result true for n = 1
1 0
Assuming that k = ( F ) + ( F ) …
k k − 1
… it follows that 𝜙𝑘 + 1 = 𝜙((𝐹 )𝜙+(𝐹 ))
𝑘 𝑘 − 1
( ) ( ) ( )
  F + 1 + F
=
k k − 1
=
( ) ( ) ( )
F +F + F = F +F
k k − 1 k k + 1 k
and result is also true for n = k + 1
Since true for n = 1 and (true for n = k  true for n = k +
AnswerMarks
1), the result follows for all n  1 by inductionB1
M1
M1
M1
A1*
A1dep
AnswerMarks
[6]2.5
1.2
1.1a
3.1a
2.2a
AnswerMarks
2.4Induction hypothesis explicitly stated (so “assume the
result is true for n = k” does not suffice without a clear
indication as to what it is that is being assumed)
k + 1 
Attempt at with explicit use of induction
hypothesis
Use of result for n = 2 or from known property of 
(e.g. from the auxiliary equation of the Fib. sequence)
or by direct calculation
(k + 1)th case established clearly from use of the
defining Fibonacci sequence property
Explanation of the inductive logic (can only be
awarded if the result has actually been proven).
Question 3:
3 | 1 = 1 . + 0 = ( F ) + ( F )    so result true for n = 1
1 0
Assuming that k = ( F ) + ( F ) …
k k − 1
… it follows that 𝜙𝑘 + 1 = 𝜙((𝐹 )𝜙+(𝐹 ))
𝑘 𝑘 − 1
( ) ( ) ( )
  F + 1 + F
=
k k − 1
=
( ) ( ) ( )
F +F + F = F +F
k k − 1 k k + 1 k
and result is also true for n = k + 1
Since true for n = 1 and (true for n = k  true for n = k +
1), the result follows for all n  1 by induction | B1
M1
M1
M1
A1*
A1dep
[6] | 2.5
1.2
1.1a
3.1a
2.2a
2.4 | Induction hypothesis explicitly stated (so “assume the
result is true for n = k” does not suffice without a clear
indication as to what it is that is being assumed)
k + 1 
Attempt at with explicit use of induction
hypothesis
Use of result for n = 2 or from known property of 
(e.g. from the auxiliary equation of the Fib. sequence)
or by direct calculation
(k + 1)th case established clearly from use of the
defining Fibonacci sequence property
Explanation of the inductive logic (can only be
awarded if the result has actually been proven).
3 The irrational number $\phi = \frac { 1 } { 2 } ( 1 + \sqrt { 5 } )$ plays a significant role in the sequence of Fibonacci numbers given by $\mathrm { F } _ { 0 } = 0 , \mathrm {~F} _ { 1 } = 1$ and $\mathrm { F } _ { \mathrm { n } + 1 } = \mathrm { F } _ { \mathrm { n } } + \mathrm { F } _ { \mathrm { n } - 1 }$ for $n \geqslant 1$.

Prove by induction that, for each positive integer $n , \phi ^ { n } = \mathrm { F } _ { \mathrm { n } } \times \phi + \mathrm { F } _ { \mathrm { n } - 1 }$.

\hfill \mbox{\textit{OCR Further Additional Pure 2022 Q3 [6]}}