CAIE Further Paper 1 2024 November — Question 1 5 marks

Exam BoardCAIE
ModuleFurther Paper 1 (Further Paper 1)
Year2024
SessionNovember
Marks5
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Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve recurrence relation formula
DifficultyModerate -0.3 This is a straightforward proof by induction with a given formula for a simple recurrence relation. The base case is trivial (u₁ = 4 = 3¹ + 1), and the inductive step requires only direct substitution into the recurrence relation and basic algebra. While induction is a Further Maths topic, this particular question is more routine than average A-level difficulty due to its mechanical nature and lack of conceptual obstacles.
Spec4.01a Mathematical induction: construct proofs
1 The sequence \(u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots\) is such that \(u _ { 1 } = 4\) and \(u _ { n + 1 } = 3 u _ { n } - 2\) for \(n \geqslant 1\).
Prove by induction that \(u _ { n } = 3 ^ { n } + 1\) for all positive integers \(n\).
Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(u_1 = 4 = 3^1 + 1\)B1 Shows base case
Assume true for \(n = k\), so \(u_k = 3^k + 1\)B1 States inductive hypothesis
Then \(u_{k+1} = 3(3^k + 1) - 2 = 3^{k+1} + 1\)M1 A1 Substitutes into recursion formula
[So true for \(n = k+1\)]. Hence by induction, \(u_n = 3^n + 1\) for all positive integers.A1 States conclusion
Total: 5 
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_1 = 4 = 3^1 + 1$ | **B1** | Shows base case |
| Assume true for $n = k$, so $u_k = 3^k + 1$ | **B1** | States inductive hypothesis |
| Then $u_{k+1} = 3(3^k + 1) - 2 = 3^{k+1} + 1$ | **M1 A1** | Substitutes into recursion formula |
| [So true for $n = k+1$]. Hence by induction, $u_n = 3^n + 1$ for all positive integers. | **A1** | States conclusion |
| **Total: 5** | | |

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1 The sequence $u _ { 1 } , u _ { 2 } , u _ { 3 } , \ldots$ is such that $u _ { 1 } = 4$ and $u _ { n + 1 } = 3 u _ { n } - 2$ for $n \geqslant 1$.\\
Prove by induction that $u _ { n } = 3 ^ { n } + 1$ for all positive integers $n$.\\

\hfill \mbox{\textit{CAIE Further Paper 1 2024 Q1 [5]}}
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Q1 5 Q2 7 Q3 10 Q4 13 Q5 9 Q6 15 Q7 16