CAIE FP1 2013 June — Question 2 5 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2013
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyModerate -0.3 This is a straightforward proof by induction with a simple divisibility statement. The base case is trivial (n=1 gives 24), and the inductive step requires only basic algebraic manipulation: 5^{2(k+1)} - 1 = 25ยท5^{2k} - 1 = 25(5^{2k} - 1) + 24, where both terms are divisible by 8. While it's a Further Maths topic, it's a standard textbook exercise requiring no novel insight, making it slightly easier than average overall.
Spec4.01a Mathematical induction: construct proofs
2 Prove by mathematical induction that \(5 ^ { 2 n } - 1\) is divisible by 8 for every positive integer \(n\).
Question 2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(P_n\): \(5^{2n}-1\) is divisible by 8. \(5^2-1=24=3\times8 \Rightarrow P_1\) is trueB1, B1 Proves base case
Assume \(P_k\) is true: \(5^{2k}-1=8\lambda\) for some \(k\). \(5^{2k+2}-1=25\cdot5^{2k}-1=24\cdot5^{2k}+5^{2k}-1 = 3\times8\cdot5^{2k}+8\lambda\)M1 States inductive hypothesis
\(\therefore P_k \Rightarrow P_{k+1}\)A1 Proves inductive step
Since \(P_1\) is true and \(P_k\Rightarrow P_{k+1}\), \(\therefore P_n\) is true for every positive integer \(n\) (by PMI)A1 States conclusion
Total: [5]
## Question 2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| $P_n$: $5^{2n}-1$ is divisible by 8. $5^2-1=24=3\times8 \Rightarrow P_1$ is true | B1, B1 | Proves base case |
| Assume $P_k$ is true: $5^{2k}-1=8\lambda$ for some $k$. $5^{2k+2}-1=25\cdot5^{2k}-1=24\cdot5^{2k}+5^{2k}-1 = 3\times8\cdot5^{2k}+8\lambda$ | M1 | States inductive hypothesis |
| $\therefore P_k \Rightarrow P_{k+1}$ | A1 | Proves inductive step |
| Since $P_1$ is true and $P_k\Rightarrow P_{k+1}$, $\therefore P_n$ is **true for every positive integer** $n$ (by PMI) | A1 | States conclusion |

**Total: [5]**

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2 Prove by mathematical induction that $5 ^ { 2 n } - 1$ is divisible by 8 for every positive integer $n$.

\hfill \mbox{\textit{CAIE FP1 2013 Q2 [5]}}
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Q1 4 Q2 5 Q3 8 Q4 8 Q5 9 Q6 9 Q7 10 Q8 10 Q9 10 Q10 13 Q11 EITHER Q11 OR