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

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.

The principle of mathematical induction

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

1. IMATH_2 is true (base case), and
2. for all positive integers $k$, $P(k) \implies P(k + 1)$ (inductive step),

then $P(n)$ is true for all positive integers $n \ge 1$.

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 $2^n > n^2$), the base case is $n = 5$. The induction proves $P(n)$ for all $n \ge 5$.

The four-part structure

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

Part 1: Base case

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

Part 2: Inductive hypothesis

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

Part 3: Inductive step

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

Part 4: Conclusion

End with "By the principle of mathematical induction, the statement holds for all positive integers $n$." (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: $\sum_{i = 1}^{n} a_i = f(n)$.

Standard technique

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

Worked example: sum of squares

Prove $\sum_{i = 1}^{n} i^2 = \frac{n(n + 1)(2 n + 1)}{6}$.

Base ($n = 1$)

LHS $= 1$. RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$. Equal.

Hypothesis

Assume $\sum_{i = 1}^{k} i^2 = \frac{k(k + 1)(2 k + 1)}{6}$.

Step

IMATH_28 by hypothesis.

Factor out $(k + 1)$: $(k + 1) \left[ \frac{k(2 k + 1)}{6} + (k + 1) \right] = (k + 1) \left[ \frac{k(2 k + 1) + 6(k + 1)}{6} \right] = (k + 1) \cdot \frac{2 k^2 + 7 k + 6}{6}$.

Factor: $2 k^2 + 7 k + 6 = (k + 2)(2 k + 3)$.

So $\sum_{i = 1}^{k + 1} i^2 = \frac{(k + 1)(k + 2)(2 k + 3)}{6}$, which is the formula at $n = k + 1$.

Conclusion: By induction, the formula holds for all positive integers.

Flavour 2: Divisibility

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

Standard technique

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

Worked example: $7^n - 1$ divisible by IMATH_43

Base ($n = 1$)

IMATH_45 , divisible by $6$. Holds.

Hypothesis

IMATH_47 for some integer $M$.

Step

IMATH_49 .

Divisible by $6$.

Conclusion: By induction, $7^n - 1$ is divisible by $6$ for all positive integers $n$.

A trickier divisibility

Prove $5^n + 3$ is divisible by $4$ for all positive integers $n$.

Base ($n = 1$)

IMATH_58 , divisible by $4$.

Hypothesis

IMATH_60 .

Step

IMATH_61 . We need to relate this to $5^k + 3$.

$5 \cdot 5^k + 3 = 5(5^k + 3) - 15 + 3 = 5(5^k + 3) - 12 = 5 \cdot 4 M - 12 = 4(5 M - 3)$.

Divisible by $4$.

Conclusion: By induction, the result holds.

Flavour 3: Inequalities

Prove $E(n) \le F(n)$ (or strict) for all $n \ge n_0$.

Standard technique

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

Worked example: $2^n > n^2$ for IMATH_72

Base ($n = 5$)

IMATH_74 . Holds.

Hypothesis

Assume $2^k > k^2$ for some $k \ge 5$.

Step

IMATH_77 by hypothesis.

We want this to be $\ge (k + 1)^2 = k^2 + 2 k + 1$. That is, $2 k^2 \ge k^2 + 2 k + 1$, that is $k^2 - 2 k - 1 \ge 0$, that is $k \ge 1 + \sqrt{2} \approx 2.41$.

For $k \ge 5 > 2.41$, this holds. So $2^{k + 1} > 2 k^2 \ge (k + 1)^2$.

Conclusion: By induction, $2^n > n^2$ for all $n \ge 5$.

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 $n = k$. The step proves the statement at $n = k + 1$. Do not phrase the step as "assume $P(k + 1)$" or substitute $P(k + 1)$ 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 $f(k + 1)$ form, write out $f(k + 1)$ explicitly first and aim for it.

Trap: Off-by-one indexing

$\sum_{i = 1}^{k + 1} a_i = \sum_{i = 1}^{k} a_i + a_{k + 1}$. The new term added is $a_{k + 1}$, not $a_k$.

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 $P(k - 1)$ in addition to $P(k)$, you need strong induction. State the hypothesis as "assume $P(j)$ holds for all $1 \le j \le k$".

Exam strategy

In Section II, induction questions typically run 4-6 marks. Allocate $5-8$ 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 $n = k + 1$ and work backwards. Markers reward partial progress.

Practice patterns

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

Prove $\sum_{i = 1}^{n} (3 i - 1) = \frac{n(3 n + 1)}{2}$, $\sum_{i = 1}^{n} 2^i = 2^{n + 1} - 2$, or $\sum_{i = 1}^{n} i(i + 1)(i + 2) = \frac{n(n + 1)(n + 2)(n + 3)}{4}$.

Pattern B: divisibility by a fixed integer

Prove $11^n - 4^n$ is divisible by $7$, or $n^3 + 5 n$ is divisible by $6$.

Pattern C: exponential beats polynomial

Prove $3^n > 2 n^2$ for $n \ge 4$, or $n! > 3^n$ for $n \ge 7$.

Pattern D: recursive sequence has a given closed form

Sequence $a_1 = 1$, $a_{n + 1} = 3 a_n + 2$. Prove $a_n = 2 \cdot 3^{n - 1} - 1$.

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.

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.