Proof | HSC Extension 1

Assessment evidence

What these examples cover

These examples show assessable skills, not an official HSC topic weighting or prediction. The 2024 NSW syllabuses have their first HSC examination in 2027; students sitting the 2026 HSC continue under the applicable 2017 syllabus.

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Teacher verification note: check the mathematics, notation, permitted technology and syllabus fit for your cohort before classroom or assessment use.

3 deterministic examples

Proof questions with worked solutions

Each question is generated from an existing curriq QDSL archetype at a fixed seed. Marks and catalog content mappings travel with the generated question.

Worked example 1

3 marksext1-p1-1

Induction and divisibility Prove by induction that

9^(2n)-1

is divisible by 8 for all integers

n>=1

Part a · 3 marks

Give a complete induction proof.

Worked solution

Answer: Verify n=1, assume the result for n=k, then write the next expression as 81 times the hypothesis plus 81-1, both divisible by 8.

Induction Base: 80
Induction Step: 81

Catalog reference: ext1_sgs25_induction_odd_square · seed 41001

Worked example 2

3 marksext1-p1-1

Induction and divisibility Prove by induction that

15^(2^n)-1

is divisible by

2^(n+2)

for every positive integer n.

Part a · 3 marks

Give a complete induction proof.

Worked solution

Answer: The base case follows because $15^2-1$ is divisible by 8. Assuming $15^(2^k)-1=2^(k+2)r$ , factor the next expression as the product of this term and $15^(2^k)+1$ , whose second factor is even.

Induction Base Case: $8$ divides $15^2-1$ .
Induction Factorisation: $A^2-1=(A-1)(A+1)$ .
Induction Conclusion: The product contains $2^(k+3)$ .

Catalog reference: ext1_jr25_induction_power_divisibility · seed 41001

Worked example 3

3 marksext1-p1-1

Induction and divisibility Use mathematical induction to prove that

3^(2n)+7

is divisible by 8 for every positive integer

n>=1

Part a · 3 marks

Write a complete induction proof.

Worked solution

Answer: Verify the base case, assume $3^(2k)+7=8m$ , and rewrite the next term as $8(9m-71)$ .

Induction Base Case: $3^2+7=16$ is divisible by 8.
Induction Hypothesis: Assume $3^(2k)+7=8m$ .
Induction Step: Substitute into $3^(2(k+1))+7$ and factor out 8.

Catalog reference: ext1_pem_2025_q12b_induction_divisibility · seed 41001

Content mapping and common errors

Proof content identifiers

ME-P1

Proof by Mathematical Induction

Understand the principle of mathematical induction; prove results for sums, divisibility and inequalities.

Common student mistakes

·Not assuming P(k) is true before proving P(k+1)
·Missing the explicit base-case check
·Incomplete inductive step — not clearly deriving the k+1 form from the k assumption

Source and methodology

Examples are deterministic curriq catalog generations from existing QDSL archetypes at the published fixed seed. The generator copies only student-facing prompts, mark allocations, topic mappings, display tables or options, checked answers and marking-path results into this resource.

Review the official NSW Curriculum source ↗

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What these examples cover

Proof questions with worked solutions

Worked example 1

Worked example 2

Worked example 3

Proof content identifiers

Proof by Mathematical Induction

Source and methodology

Other Extension 1 topics

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