Why do you need a base case in mathematical induction?

The base case starts the chain. Without it, the inductive step only says 'if it is true at k then it is true at k + 1', but nothing ever makes it start being true. A false statement can even have a valid inductive step, so both the base case and the inductive step are required.

Why do you need a base case in mathematical induction?

The base case starts the chain. Without it, the inductive step only says 'if it is true at k then it is true at k + 1', but nothing ever makes it start being true. A false statement can even have a valid inductive step, so both the base case and the inductive step are required.

Year 12 Maths Extension 1 (2027) Proof

Proof of sums by induction

Q: What is proof by mathematical induction?

Mathematical induction is a method for proving a statement is true for every integer n from a starting value onwards. You prove a base case (the statement holds at the start), then prove an inductive step (if it holds at n = k, it also holds at n = k + 1). Together these force the statement to be true for all n.

Q: What are the steps of a proof by induction?

There are three steps. Step 1, the base case: show the statement is true for the smallest value, usually n = 1. Step 2, the inductive hypothesis: assume the statement is true for some integer n = k. Step 3, the inductive step: use that assumption to prove the statement for n = k + 1. Then conclude by the principle of mathematical induction.

Q: How do you prove a sum formula by induction?

Check the base case by substituting the smallest n. Assume the sum formula holds for n = k. Then add the next term (the k plus 1 term) to both sides of the assumed formula and simplify until you reach the formula with n = k + 1. Finish with a concluding sentence citing the principle of induction.

Q: What is the inductive hypothesis?

The inductive hypothesis is the assumption that the statement is true for some particular integer n = k. It is an assumption for one value of k, not for all n. You are allowed to use it inside the inductive step to prove the n = k + 1 case.

20 practice questions 2 video lessons Theory + worked examples

Learn proof by mathematical induction for series in NSW Year 12 Mathematics Extension 1. Induction confirms a closed-form formula for a sum holds for every positive integer by establishing a base case, then showing each case forces the next.

You will learn to set out the base case, assume the inductive hypothesis and prove the step to the next term, then write the concluding statement examiners expect — the key technique for proving summation identities in the HSC Extension 1 course.

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Practice questions

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Proof by Induction | Proving a Series by Induction Watch
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Theory

Proof by mathematical induction is a NSW Year 12 Mathematics Extension 1 focus area (NESA outcome ME1-12-01). This page shows how to prove a sum formula by induction using the base case and the inductive step, with four fully worked examples.

Mathematical induction is a method for proving that a statement $S (n)$ is true for every integer $n$ from a starting value $n_{0}$ onwards. The intuition is a row of dominoes: if you knock over the first, and each falling domino knocks over the next, then every domino falls.

A proof by induction has three parts. The base case shows the statement is true at the start, usually $S (1)$ . The inductive hypothesis assumes the statement is true for some integer $n = k$ . The inductive step uses that assumption to prove the statement for the next value, $n = k + 1$ .

For a sum, the statement $S (n)$ is a formula for a running total, such as $1 + 2 + 3 + \dots + n = \frac{n (n + 1)}{2}$ . The inductive step works by adding the next term to both sides of the assumed formula and showing the result matches the formula with $n$ replaced by $k + 1$ .

NESA link. This topic sits in the Year 12 Proof by mathematical induction focus area. Outcome ME1-12-01 — "uses mathematical induction to prove results involving sums and divisibility" — together with the working-mathematically outcome MAO-WM-01. The syllabus asks students to examine the nature of inductive proof, including the statement to be proved, the base case and the inductive step, and to prove results for sums using mathematical induction.

The chain of reasoning: prove the first case, then that each case forces the next.

The base case tips the first domino; the inductive step makes each one topple the next.

To prove $S (n)$ for all integers $n \geq n_{0}$ , establish two things:

Base case: S (n_{0}) is true.

Inductive step: S (k) \Rightarrow S (k + 1) for all k \geq n_{0} .

S (k) \Rightarrow S (k + 1)

For a sum, if the inductive hypothesis is

t_{1} + t_{2} + \dots + t_{k} = f (k),

then the inductive step adds the next term $t_{k + 1}$ to both sides and shows

f (k) + t_{k + 1} = f (k + 1) .

f (k) + t_{k + 1} = f (k + 1)

Key principle. The base case and the inductive step are both required. The base case starts the chain; the inductive step continues it. Prove only one and the statement need not be true.

Standard sum results (worth knowing)

Sum	Closed form
$1 + 2 + 3 + \dots + n$	$\frac{n (n + 1)}{2}$
$1 + 3 + 5 + \dots + (2 n - 1)$	$n^{2}$
$1^{2} + 2^{2} + 3^{2} + \dots + n^{2}$	$\frac{n (n + 1) (2 n + 1)}{6}$
$1 + 2 + 2^{2} + \dots + 2^{n}$	$2^{n + 1} - 1$

The second and fourth rows are the two sum examples named in the NESA Extension 1 syllabus (ME1-12-01).

How to prove a sum formula by induction

State and test the base case. Write the statement $S (n)$ , then substitute the smallest value (usually $n = 1$ ) and check the left-hand side equals the right-hand side.
Assume the hypothesis. Assume $S (k)$ is true for some integer $k \geq n_{0}$ — this is the inductive hypothesis. Write the assumed formula out in full.
Prove the step. Add the $(k + 1)$ th term to both sides of the assumed formula. Simplify the right-hand side by factoring until it matches the original formula with $n$ replaced by $k + 1$ .
Conclude. State that, since the base case holds and the step holds, by the principle of mathematical induction $S (n)$ is true for all $n \geq n_{0}$ .

Example 1 — An arithmetic sum

Prove

3 + 7 + 11 + \dots + (4 n - 1) = n (2 n + 1)

for all

n \geq 1

Solution

Base case $(n = 1)$ : LHS $= 3$ ; RHS $= 1 \cdot 3 = 3$ . True.

Assume true for $n = k$ : $3 + 7 + \dots + (4 k - 1) = k (2 k + 1)$ .

Inductive step: add the next term $(4 k + 3)$ to both sides:

$k (2 k + 1) + (4 k + 3)$	$=$	$2 k^{2} + 5 k + 3$
	$=$	$(k + 1) (2 k + 3)$
	$=$	$(k + 1) (2 (k + 1) + 1)$

k (2 k + 1) + (4 k + 3) = (k + 1) (2 k + 3)

This is the formula at $n = k + 1$ . By the principle of induction, it holds for all $n \geq 1$ .

Example 2 — A geometric sum

Prove

5 + 5^{2} + \dots + 5^{n} = \frac{5^{n + 1} - 5}{4}

for all

n \geq 1

Solution

Base case $(n = 1)$ : LHS $= 5$ ; RHS $= \frac{25 - 5}{4} = 5$ . True.

Assume true for $n = k$ : $5 + \dots + 5^{k} = \frac{5^{k + 1} - 5}{4}$ .

Inductive step: add $5^{k + 1}$ to both sides:

$\frac{5^{k + 1} - 5}{4} + 5^{k + 1}$	$=$	$\frac{5^{k + 1} - 5 + 4 \cdot 5^{k + 1}}{4}$
	$=$	$\frac{5 \cdot 5^{k + 1} - 5}{4}$
	$=$	$\frac{5^{k + 2} - 5}{4}$

\frac{5^{k + 1} - 5}{4} + 5^{k + 1} = \frac{5^{k + 2} - 5}{4}

This is the formula at $n = k + 1$ . By the principle of induction, it holds for all $n \geq 1$ .

Example 3 — Sum of squares

Prove

1^{2} + 2^{2} + \dots + n^{2} = \frac{n (n + 1) (2 n + 1)}{6}

for all

n \geq 1

Solution

Base case $(n = 1)$ : LHS $= 1$ ; RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$ . True.

Assume true for $n = k$ : $1^{2} + \dots + k^{2} = \frac{k (k + 1) (2 k + 1)}{6}$ .

Inductive step: add $(k + 1)^{2}$ to both sides and factor out $(k + 1)$ :

$\frac{k (k + 1) (2 k + 1)}{6} + (k + 1)^{2}$	$=$	$\frac{(k + 1) [k (2 k + 1) + 6 (k + 1)]}{6}$
	$=$	$\frac{(k + 1) (2 k^{2} + 7 k + 6)}{6}$
	$=$	$\frac{(k + 1) (k + 2) (2 k + 3)}{6}$

\frac{k (k + 1) (2 k + 1)}{6} + {(k + 1)}^{2} = \frac{(k + 1) (k + 2) (2 k + 3)}{6}

This is the formula at $n = k + 1$ (since $2 (k + 1) + 1 = 2 k + 3$ ). By induction, it holds for all $n \geq 1$ .

Example 4 — NESA example (base case n = 0)

Prove

1 + 2 + 2^{2} + \dots + 2^{n} = 2^{n + 1} - 1

for all integers

n \geq 0

. (This is one of the syllabus's own examples for ME1-12-01.)

Solution

Base case $(n = 0)$ : LHS $= 2^{0} = 1$ ; RHS $= 2^{1} - 1 = 1$ . True. (Here the start value is $n_{0} = 0$ , not $1$ .)

Assume true for $n = k$ : $1 + 2 + \dots + 2^{k} = 2^{k + 1} - 1$ .

Inductive step: add the next term $2^{k + 1}$ to both sides:

$(2^{k + 1} - 1) + 2^{k + 1}$	$=$	$2 \cdot 2^{k + 1} - 1$
	$=$	$2^{k + 2} - 1$

(2^{k + 1} - 1) + 2^{k + 1} = 2^{k + 2} - 1

This is the formula at $n = k + 1$ . By the principle of induction, it holds for all $n \geq 0$ .

Common pitfalls

Skipping a step. The base case and the inductive step are both essential. A false statement can have a perfectly valid inductive step, so proving the step alone proves nothing — and checking a few values is not the base case for all

n

The wrong $(k + 1)$ th term. When you extend the sum, add the term at

n = k + 1

, not at

n = k

. For

2 + 4 + 6 + \dots + 2 n

the next term is

2 (k + 1) = 2 k + 2

, not

2 k

Not using the assumption. The inductive step must use the hypothesis

S (k)

— substitute it in. Re-deriving the whole result from scratch is not an inductive proof.

No concluding sentence. Finish by explicitly stating that, since the base case and the step both hold, the result is true for all

n \geq n_{0}

by the principle of mathematical induction.

Frequently asked questions

What is proof by mathematical induction?

It is a method for proving a statement $S (n)$ is true for every integer $n$ from a starting value onwards. You prove a base case, then prove that truth at $n = k$ forces truth at $n = k + 1$ ; together these make the statement true for all $n$ .

What are the steps of a proof by induction?

Three steps: (1) the base case — show $S (n_{0})$ is true, usually $n_{0} = 1$ ; (2) the inductive hypothesis — assume $S (k)$ is true; (3) the inductive step — use that to prove $S (k + 1)$ . Then conclude by the principle of induction.

Why do you need a base case in induction?

The base case starts the chain. Without it, the inductive step only says "if true at $k$ , then true at $k + 1$ " but nothing ever makes it start. A false statement can even have a valid step, so the base case is what pins the argument to reality.

How do you prove a sum formula by induction?

Test the base case, assume the formula at $n = k$ , then add the $(k + 1)$ th term to both sides and simplify until you reach the same formula with $n = k + 1$ . Finish with the concluding sentence.

What is the inductive hypothesis?

It is the assumption that $S (k)$ is true for some particular integer $k$ — a single value, not all $n$ . You are allowed to use it inside the inductive step to prove the $n = k + 1$ case.

Is mathematical induction in the NSW Year 12 Extension 1 course?

Yes. Proof by mathematical induction is a Year 12 focus area of Mathematics Extension 1. NESA outcome ME1-12-01 states that a student "uses mathematical induction to prove results involving sums and divisibility".

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Proof of divisibility by induction

Proof of sums by induction

📖 Theory

Standard sum results (worth knowing)

How to prove a sum formula by induction

Common pitfalls

Frequently asked questions

Theory