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HSC Maths Extension 1 mathematical induction: deep dive (2026 guide)

A complete deep dive into mathematical induction for HSC Mathematics Extension 1. The principle, the four-part structure, fully worked walkthroughs of all three flavours (series, divisibility, inequalities), the standard exam traps, and an embedded practice question set with full solutions.

Reviewed by: AI editorial process; not yet individually human-reviewed

Jump to a section
  1. What this guide covers
  2. The principle of mathematical induction
  3. The four-part structure
  4. Flavour 1: Series identities
  5. Flavour 2: Divisibility
  6. Flavour 3: Inequalities
  7. Flavour 4: Recursive sequences with a closed form
  8. Common traps and how to avoid them
  9. Exam strategy
  10. Practice patterns
  11. Where induction appears in the wider syllabus
  12. Check your knowledge
  13. Linked dot points

What this guide covers

Mathematical induction is one of the highest-yield topics in HSC Mathematics Extension 1. It appears nearly every year, the structure is rigid, and the algebra is standardised. With practice, you should be able to bank these marks reliably.

This guide walks through the four-part structure, the three standard flavours (series, divisibility, inequalities), and the common traps. By the end you should have a template you can deploy on any HSC induction question, plus a set of practice questions with full step-by-step solutions.

The principle of mathematical induction

Let P(n)P(n) be a statement about a positive integer nn. If:

  1. P(1)P(1) is true (base case), and
  2. for all positive integers kk, P(k)β€…β€ŠβŸΉβ€…β€ŠP(k+1)P(k) \implies P(k + 1) (inductive step),

then P(n)P(n) is true for all positive integers nβ‰₯1n \ge 1.

The intuition is a chain of dominoes. The base case knocks down the first; the inductive step ensures each falling domino knocks down the next. So every domino falls.

The starting integer can be anything. For some statements (like 2n>n22^n > n^2), the base case is n=5n = 5. The induction proves P(n)P(n) for all nβ‰₯5n \ge 5.

The four-part structure

Every induction proof has exactly four parts. Markers want to see each one labelled.

Part 1: Base case

Substitute n=1n = 1 (or the smallest valid nn) into the statement and verify it by direct calculation. Show your working.

Part 2: Inductive hypothesis

Write "Assume the statement holds for n=kn = k, that is, [state the statement at n=kn = k]". This is the assumption you will use in the next part.

Part 3: Inductive step

Show that the statement holds at n=k+1n = k + 1. The technique depends on the flavour (series, divisibility, inequality), but the core move is to express the n=k+1n = k + 1 form in terms of the n=kn = k form, then substitute the hypothesis.

Part 4: Conclusion

End with "By the principle of mathematical induction, the statement holds for all positive integers nn." (Or the appropriate range.)

If a conclusion sentence is missing, markers can deduct a mark, even with otherwise correct working.

Flavour 1: Series identities

The most common HSC induction is a sum identity: βˆ‘i=1nai=f(n)\sum_{i = 1}^{n} a_i = f(n).

Standard technique

In the inductive step, write βˆ‘i=1k+1ai=βˆ‘i=1kai+ak+1\sum_{i = 1}^{k + 1} a_i = \sum_{i = 1}^{k} a_i + a_{k + 1}, substitute the hypothesis for the kk-term sum, then simplify to match f(k+1)f(k + 1).

Flavour 2: Divisibility

Prove that E(n)E(n) is divisible by dd for all positive integers.

Standard technique

In the inductive step, manipulate E(k+1)E(k + 1) to expose a multiple of dd. Often the trick is to write E(k+1)E(k + 1) in terms of E(k)E(k) (using the hypothesis E(k)=dME(k) = d M) plus a remainder that is also a multiple of dd.

Flavour 3: Inequalities

Prove E(n)≀F(n)E(n) \le F(n) (or strict) for all nβ‰₯n0n \ge n_0.

Standard technique

In the inductive step, show that the growth from n=kn = k to n=k+1n = k + 1 on one side is at least as much as on the other. Use the hypothesis to bound E(k)E(k) in terms of F(k)F(k), then extend.

Flavour 4: Recursive sequences with a closed form

A recursively defined sequence a1a_1, an+1=g(an)a_{n+1} = g(a_n) is often given with a proposed closed form. Prove the closed form by induction.

Common traps and how to avoid them

Trap: Skipping the base case

A base case is not optional. Without it, your inductive step has nothing to start from. Always include it explicitly.

Trap: Assuming the conclusion

The hypothesis is the statement at n=kn = k. The step proves the statement at n=k+1n = k + 1. Do not phrase the step as "assume P(k+1)P(k + 1)" or substitute P(k+1)P(k + 1) into your work as a known fact.

Trap: Algebra in the step

Most exam errors are algebraic, not logical. Factor out common terms early. If you cannot match the target f(k+1)f(k + 1) form, write out f(k+1)f(k + 1) explicitly first and aim for it.

Trap: Off-by-one indexing

βˆ‘i=1k+1ai=βˆ‘i=1kai+ak+1\sum_{i = 1}^{k + 1} a_i = \sum_{i = 1}^{k} a_i + a_{k + 1}. The new term added is ak+1a_{k + 1}, not aka_k.

Trap: Missing conclusion

Mark-up rubrics often allocate a mark for the explicit "by the principle of mathematical induction" sentence. Do not forget it.

Trap: Strong induction without saying

If your step uses P(kβˆ’1)P(k - 1) in addition to P(k)P(k), you need strong induction. State the hypothesis as "assume P(j)P(j) holds for all 1≀j≀k1 \le j \le k".

Exam strategy

In Section II, induction questions typically run 4-6 marks. Allocate 5βˆ’85-8 minutes per question. The proof has four parts, each worth approximately one mark:

  • Base case (1 mark)
  • Inductive hypothesis (1 mark)
  • Inductive step, including correct algebraic manipulation and explicit use of the hypothesis (2-3 marks)
  • Conclusion (1 mark)

If you get stuck on the algebra, write down what the formula should look like at n=k+1n = k + 1 and work backwards. Markers reward partial progress.

Practice patterns

Pattern A: sum of an arithmetic-like or geometric-like series

Prove βˆ‘i=1n(3iβˆ’1)=n(3n+1)2\sum_{i = 1}^{n} (3 i - 1) = \frac{n(3 n + 1)}{2}, βˆ‘i=1n2i=2n+1βˆ’2\sum_{i = 1}^{n} 2^i = 2^{n + 1} - 2, or βˆ‘i=1ni(i+1)(i+2)=n(n+1)(n+2)(n+3)4\sum_{i = 1}^{n} i(i + 1)(i + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}.

Pattern B: divisibility by a fixed integer

Prove 11nβˆ’4n11^n - 4^n is divisible by 77, or n3+5nn^3 + 5 n is divisible by 66.

Pattern C: exponential beats polynomial

Prove 3n>2n23^n > 2 n^2 for nβ‰₯4n \ge 4, or n!>3nn! > 3^n for nβ‰₯7n \ge 7.

Pattern D: recursive sequence has a given closed form

Sequence a1=1a_1 = 1, an+1=3an+2a_{n + 1} = 3 a_n + 2. Prove an=2β‹…3nβˆ’1βˆ’1a_n = 2 \cdot 3^{n - 1} - 1.

Where induction appears in the wider syllabus

Induction is not just a stand-alone topic. It underpins:

  • Proof of Pascal's rule and the binomial theorem.
  • Verification of closed-form formulas for recursively defined sequences.
  • Proofs of divisibility results that show up in number theory (Extension 2).
  • Geometric proofs by induction on the number of sides, regions, or vertices.

The technique is genuinely useful for the rest of your maths education, not just an HSC topic.

Check your knowledge

Try these questions before reading the solutions. They are roughly arranged from easier to harder.

  1. Prove by induction that βˆ‘i=1n(2iβˆ’1)=n2\sum_{i=1}^{n} (2i - 1) = n^2 for all integers nβ‰₯1n \geq 1.
  2. Prove by induction that βˆ‘i=1n2i=2n+1βˆ’2\sum_{i=1}^{n} 2^i = 2^{n+1} - 2 for all integers nβ‰₯1n \geq 1.
  3. Prove by induction that 9nβˆ’19^n - 1 is divisible by 88 for all integers nβ‰₯1n \geq 1.
  4. Prove by induction that βˆ‘i=1niβ‹…i!=(n+1)!βˆ’1\sum_{i=1}^{n} i \cdot i! = (n+1)! - 1 for all integers nβ‰₯1n \geq 1.
  5. Prove by induction that 11nβˆ’4n11^n - 4^n is divisible by 77 for all integers nβ‰₯1n \geq 1.
  6. Prove by induction that 2nβ‰₯n+12^n \geq n + 1 for all integers nβ‰₯1n \geq 1.
  7. Prove by induction that n!>2nn! > 2^n for all integers nβ‰₯4n \geq 4.
  8. The sequence is defined by a1=2a_1 = 2 and an+1=2an+1a_{n+1} = 2 a_n + 1 for nβ‰₯1n \geq 1. Prove by induction that an=3β‹…2nβˆ’1βˆ’1a_n = 3 \cdot 2^{n-1} - 1 for all integers nβ‰₯1n \geq 1.

Linked dot points

For more detail on the three flavours, see the dot point pages at induction on series, induction on divisibility, induction on inequalities, and induction on general statements.

For NESA past papers and marking guidelines, refer to educationstandards.nsw.edu.au.

  • induction
  • proof
  • hsc-maths-extension-1
  • year-12
  • 2026
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