CAIE FP1 2018 June — Question 2 6 marks

Exam BoardCAIE
ModuleFP1 (Further Pure Mathematics 1)
Year2018
SessionJune
Marks6
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Mark schemeDownload PDF ↗
TopicProof by induction
TypeProve divisibility
DifficultyStandard +0.8 This is a non-standard induction proof requiring algebraic insight to simplify f(k+1) - f(k) or f(k) + f(k+1) to reveal the divisibility by 9. The hint guides students toward the key manipulation (showing f(k) + f(k+1) is divisible by 9), but executing this requires careful index manipulation with powers of 2 and 8, then factoring appropriately. More challenging than routine induction proofs of summation formulas, but still within reach for Further Maths students who recognize the algebraic structure.
Spec1.02a Indices: laws of indices for rational exponents4.01a Mathematical induction: construct proofs
2 It is given that \(\mathrm { f } ( n ) = 2 ^ { 3 n } + 8 ^ { n - 1 }\). By simplifying \(\mathrm { f } ( k ) + \mathrm { f } ( k + 1 )\), or otherwise, prove by mathematical induction that \(\mathrm { f } ( n )\) is divisible by 9 for every positive integer \(n\).
Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
We have that \(f(1) = 9\) is divisible by 9.B1 Checks base case.
Assume that \(f(k)\) is divisible by 9.B1 Makes general statement.
\(f(k+1) + f(k) = 2^{3k+3} + 8^k + 2^{3k} + 8^{k-1}\)M1 Uses expansion of \(f(k+1)\).
\(2^3 \cdot 2^{3k} + 8 \cdot 8^{k-1} + 2^{3k} + 8^{k-1}\) OEA1 Correct split of powers
\(= 9\!\left(2^{3k} + 8^{k-1}\right)\) OE so \(f(k+1)\) is divisible by 9.A1 Alt method: \(f(k+1) = 2^{3k+3} + 8^k\) \(= 2^3 \cdot 2^{3k} + 8 \cdot 8^{k-1}\) M1 \(= 8f(k)\) A1
So if \(f(k)\) is divisible by 9, so is \(f(k+1)\), (and \(f(1)\) is divisible by 9), \(f(n)\) is divisible by 9 for every integer \(n \geqslant 1\)A1
Total: 6 
**Question 2:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| We have that $f(1) = 9$ is divisible by 9. | B1 | Checks base case. |
| Assume that $f(k)$ is divisible by 9. | B1 | Makes general statement. |
| $f(k+1) + f(k) = 2^{3k+3} + 8^k + 2^{3k} + 8^{k-1}$ | M1 | Uses expansion of $f(k+1)$. |
| $2^3 \cdot 2^{3k} + 8 \cdot 8^{k-1} + 2^{3k} + 8^{k-1}$ OE | A1 | Correct split of powers |
| $= 9\!\left(2^{3k} + 8^{k-1}\right)$ OE so $f(k+1)$ is divisible by 9. | A1 | Alt method: $f(k+1) = 2^{3k+3} + 8^k$ $= 2^3 \cdot 2^{3k} + 8 \cdot 8^{k-1}$ M1 $= 8f(k)$ A1 |
| So if $f(k)$ is divisible by 9, so is $f(k+1)$, (and $f(1)$ is divisible by 9), $f(n)$ is divisible by 9 for every integer $n \geqslant 1$ | A1 | |
| **Total: 6** | | |
2 It is given that $\mathrm { f } ( n ) = 2 ^ { 3 n } + 8 ^ { n - 1 }$. By simplifying $\mathrm { f } ( k ) + \mathrm { f } ( k + 1 )$, or otherwise, prove by mathematical induction that $\mathrm { f } ( n )$ is divisible by 9 for every positive integer $n$.\\

\hfill \mbox{\textit{CAIE FP1 2018 Q2 [6]}}
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Q1 5 Q2 6 Q3 8 Q4 8 Q5 8 Q6 9 Q7 10 Q8 10 Q9 10 Q10 12 Q11 EITHER Q11 OR