NSWMaths Extension 1Syllabus dot point

How do we use mathematical induction to prove identities involving sums of a sequence?

Prove identities for sums of series using the principle of mathematical induction

A focused answer to the HSC Maths Extension 1 dot point on induction proofs of series identities. The base case, induction hypothesis, induction step, and conclusion, applied to sums of integers, squares, cubes and geometric-like patterns, with worked examples.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to use mathematical induction to prove that an identity for a finite sum holds for every positive integer $n$ . The proof has a standard four-part structure that you must reproduce exactly.

The answer

The principle

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

IMATH_9 is true (base case), and
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 base case may sometimes be $n = 0$ or some other starting integer, depending on what the problem asserts.)

The standard four-part structure

Part 1: Base case: Verify $P(1)$ directly by substituting $n = 1$ into both sides.
Part 2: Inductive hypothesis: Assume $P(k)$ for some positive integer $k$ . Write the assumed identity explicitly.
Part 3: Inductive step: Use the hypothesis to derive $P(k + 1)$ . The standard technique is to write the $(k + 1)$ -term sum as the $k$ -term sum (which the hypothesis gives a formula for) plus the new $(k + 1)$ -th term, then simplify to the right form.
Part 4: Conclusion: By the principle of mathematical induction, $P(n)$ holds for all positive integers $n$ .

Common series formulas (good to memorise)

\sum_{i = 1}^{n} i = \frac{n(n + 1)}{2},

\sum_{i = 1}^{n} i^2 = \frac{n(n + 1)(2 n + 1)}{6},

\sum_{i = 1}^{n} i^3 = \left( \frac{n(n + 1)}{2} \right)^2,

\sum_{i = 0}^{n - 1} a r^i = a \cdot \frac{r^n - 1}{r - 1} \quad (r \neq 1).

You should be able to prove each of these by induction.

The induction-step algebra

The hardest part is the algebra in the inductive step. The pattern:

Start with LHS for $n = k + 1$ . Split off the last term.
Substitute the hypothesis for the $k$ -term sum.
Factor or simplify to match the formula for $n = k + 1$ .

Always check that your simplified expression really does equal the RHS at $n = k + 1$ . If you get stuck, write both sides for $n = k + 1$ first and aim for the target.

Worked example

Induction on a quadratic-coefficient sum

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

Base ( $n = 1$ ): LHS $= 1$ , RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$ . Holds.
Hypothesis: IMATH_4 .
Step: IMATH_5 .

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

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

So $\sum_{i = 1}^{k + 1} i^2 = \frac{(k + 1)(k + 2)(2 k + 3)}{6} = \frac{(k + 1)((k + 1) + 1)(2(k + 1) + 1)}{6}$ . Matches.

Conclusion: by induction, the formula holds for all positive integers $n$ .

Geometric series

Prove $1 + r + r^2 + \dots + r^{n - 1} = \frac{r^n - 1}{r - 1}$ for $r \neq 1$ .

Base ( $n = 1$ ): LHS $= 1$ , RHS $= \frac{r - 1}{r - 1} = 1$ . Holds.
Hypothesis: IMATH_16 .
Step: IMATH_17 .

This matches the formula at $n = k + 1$ . By induction, the result holds for all positive integers $n$ .

Sum of cubes

Prove $\sum_{i = 1}^{n} i^3 = \left( \frac{n(n + 1)}{2} \right)^2$ .

Base ( $n = 1$ ): LHS $= 1$ , RHS $= \left( \frac{1 \cdot 2}{2} \right)^2 = 1$ . Holds.
Hypothesis: IMATH_24 .
Step: IMATH_25 .

Matches the formula at $n = k + 1$ .

A non-standard sum

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

Base ( $n = 1$ ): LHS $= 2$ , RHS $= \frac{1 \cdot 2 \cdot 3}{3} = 2$ . Holds.
Hypothesis: IMATH_31 .
Step: IMATH_32 . Matches.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2023 HSC Q155 marksProve by mathematical induction that for all positive integers

n

1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2}

Show worked answer →

Base case ( $n = 1$ ): LHS $= 1$ , RHS $= \frac{1 \cdot 2}{2} = 1$ . Equal, so the statement holds for $n = 1$ .

Inductive hypothesis: Assume the statement holds for $n = k$ , that is

$1 + 2 + \dots + k = \frac{k(k + 1)}{2}$ .

Inductive step: Show the statement holds for $n = k + 1$ :

$1 + 2 + \dots + k + (k + 1) = \frac{k(k + 1)}{2} + (k + 1)$ by the hypothesis.

$= (k + 1) \left[ \frac{k}{2} + 1 \right] = (k + 1) \cdot \frac{k + 2}{2} = \frac{(k + 1)(k + 2)}{2}$ .

This is the formula with $n$ replaced by $k + 1$ . So the statement holds for $n = k + 1$ whenever it holds for $n = k$ .

Conclusion: By the principle of mathematical induction, the statement holds for all positive integers $n$ .

Markers reward each of the four parts (base, hypothesis, step, conclusion), the algebraic manipulation in the step, and the explicit citation of the inductive hypothesis.