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

HSCMaths Extension 1Study guide22 min readUpdated 25 May 2026

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.

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What this guide covers
The principle of mathematical induction
The four-part structure
Flavour 1: Series identities
Flavour 2: Divisibility
Flavour 3: Inequalities
Flavour 4: Recursive sequences with a closed form
Common traps and how to avoid them
Exam strategy
Practice patterns
Where induction appears in the wider syllabus
Check your knowledge
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 $P(n)$ be a statement about a positive integer $n$ . If:

$P(1)$ 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 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 $f(k + 1)$ .

Prove by induction that

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

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ). Verify the statement at the smallest valid integer. LHS $= 1^2 = 1$ . RHS $= \frac{1 \cdot 2 \cdot 3}{6} = 1$ . LHS $=$ RHS, so the statement holds for $n = 1$ .

Step 2. Inductive hypothesis. Assume the statement holds for some integer $n = k \geq 1$ , that is,

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

Step 3. Set up the $n = k + 1$ sum. Split off the last term so the hypothesis can be substituted directly.

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

Step 4. Common factor. Factor out $\frac{k+1}{6}$ so the algebra collapses to a single product.

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

Step 5. Factor the quadratic. Confirm the expression matches the formula at $n = k + 1$ . Since $2k^2 + 7k + 6 = (k + 2)(2k + 3)$ ,

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

This is the formula at $n = k + 1$ , so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $\sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6}$ for all integers $n \geq 1$ .

Prove by induction that

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

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ). LHS $= 1 \cdot 2 = 2$ . RHS $= \frac{1 \cdot 2 \cdot 3}{3} = 2$ . The statement holds for $n = 1$ .

Step 2. Inductive hypothesis. Assume the statement holds at $n = k$ :

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

Step 3. Split off the $(k+1)$ -th term. Use the hypothesis to rewrite the partial sum.

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

Step 4. Common factor $\frac{(k+1)(k+2)}{3}$ . Pull it out and combine the bracket.

= \frac{(k+1)(k+2) \left[ k + 3 \right]}{3} = \frac{(k+1)(k+2)(k+3)}{3}.

Step 5. Match the target. The right-hand side at $n = k + 1$ is $\frac{(k+1)\big((k+1)+1\big)\big((k+1)+2\big)}{3} = \frac{(k+1)(k+2)(k+3)}{3}$ . The two expressions agree, so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $\sum_{i=1}^{n} i(i+1) = \frac{n(n+1)(n+2)}{3}$ for all integers $n \geq 1$ .

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 $d$ .

Prove by induction that

7^n - 1

is divisible by

6

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ): $7^1 - 1 = 6 = 6 \cdot 1$ , which is divisible by $6$ . The statement holds for $n = 1$ .
Step 2. Inductive hypothesis: Assume the statement holds at $n = k$ , that is, $7^k - 1 = 6M$ for some integer $M$ . Equivalently, $7^k = 6M + 1$ .
Step 3. Rewrite the $n = k + 1$ expression: Pull out one factor of $7$ so the hypothesis can be substituted.

7^{k+1} - 1 = 7 \cdot 7^k - 1 = 7(6M + 1) - 1.

Step 4. Expose a multiple of $6$ . Expand and simplify.

7(6M + 1) - 1 = 42M + 7 - 1 = 42M + 6 = 6(7M + 1).

Since $7M + 1$ is an integer, $7^{k+1} - 1$ is divisible by $6$ , so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $7^n - 1$ is divisible by $6$ for all integers $n \geq 1$ .

Prove by induction that

5^n + 3

is divisible by

4

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ): $5^1 + 3 = 8 = 4 \cdot 2$ , which is divisible by $4$ . The statement holds for $n = 1$ .
Step 2. Inductive hypothesis: Assume the statement holds at $n = k$ , that is, $5^k + 3 = 4M$ for some integer $M$ . Equivalently, $5^k = 4M - 3$ .
Step 3. Rewrite the $n = k + 1$ expression: Pull out one factor of $5$ .

5^{k+1} + 3 = 5 \cdot 5^k + 3 = 5(4M - 3) + 3.

Step 4. Simplify and factor out $4$ .

5(4M - 3) + 3 = 20M - 15 + 3 = 20M - 12 = 4(5M - 3).

Since $5M - 3$ is an integer, $5^{k+1} + 3$ is divisible by $4$ , so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $5^n + 3$ is divisible by $4$ for all integers $n \geq 1$ .

Prove by induction that

n^3 + 5n

is divisible by

6

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ): $1^3 + 5 \cdot 1 = 6 = 6 \cdot 1$ , which is divisible by $6$ . The statement holds for $n = 1$ .
Step 2. Inductive hypothesis: Assume the statement holds at $n = k$ , that is, $k^3 + 5k = 6M$ for some integer $M$ .
Step 3. Expand $(k+1)^3 + 5(k+1)$: Use the binomial expansion so the hypothesis can be substituted.

(k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5 = (k^3 + 5k) + 3k^2 + 3k + 6.

Step 4. Substitute the hypothesis and factor. Replace $k^3 + 5k$ with $6M$ and tidy.

(k^3 + 5k) + 3k^2 + 3k + 6 = 6M + 3k(k+1) + 6.

Step 5. Show $3k(k+1)$ is divisible by $6$ . Among any two consecutive integers $k$ and $k + 1$ , one is even, so $k(k+1)$ is even. Write $k(k+1) = 2N$ for some integer $N$ . Then $3k(k+1) = 6N$ , and

(k+1)^3 + 5(k+1) = 6M + 6N + 6 = 6(M + N + 1),

a multiple of $6$ . So $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $n^3 + 5n$ is divisible by $6$ for all integers $n \geq 1$ .

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.

Prove by induction that

2^n > n^2

for all integers

n \geq 5

Step 1. Base case ( $n = 5$ ): $2^5 = 32$ and $5^2 = 25$ . Since $32 > 25$ , the statement holds for $n = 5$ .
Step 2. Inductive hypothesis: Assume the statement holds at some integer $n = k \geq 5$ , that is, $2^k > k^2$ .
Step 3. Set up the $n = k + 1$ inequality: Use the hypothesis after doubling.

2^{k+1} = 2 \cdot 2^k > 2k^2.

Step 4. Show $2k^2 \geq (k+1)^2$ for $k \geq 5$ . Expand the target: $(k+1)^2 = k^2 + 2k + 1$ . Compare:

2k^2 - (k^2 + 2k + 1) = k^2 - 2k - 1.

The roots of $k^2 - 2k - 1 = 0$ are $k = 1 \pm \sqrt{2}$ , so $k^2 - 2k - 1 \geq 0$ when $k \geq 1 + \sqrt{2} \approx 2.41$ . Since $k \geq 5 > 1 + \sqrt{2}$ , we have $2k^2 \geq (k+1)^2$ .

Step 5. Combine. Chain the inequalities:

2^{k+1} > 2k^2 \geq (k+1)^2.

So $2^{k+1} > (k+1)^2$ and $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $2^n > n^2$ for all integers $n \geq 5$ .

Prove by induction that

3^n > n^3

for all integers

n \geq 4

Step 1. Base case ( $n = 4$ ): $3^4 = 81$ and $4^3 = 64$ . Since $81 > 64$ , the statement holds for $n = 4$ .
Step 2. Inductive hypothesis: Assume the statement holds at some integer $n = k \geq 4$ , that is, $3^k > k^3$ .
Step 3. Set up the $n = k + 1$ inequality: Multiply the hypothesis by $3$ .

3^{k+1} = 3 \cdot 3^k > 3k^3.

Step 4. Show $3k^3 \geq (k+1)^3$ for $k \geq 4$ . Expand $(k+1)^3 = k^3 + 3k^2 + 3k + 1$ and compare:

3k^3 - (k+1)^3 = 2k^3 - 3k^2 - 3k - 1.

At $k = 4$ : $2 \cdot 64 - 3 \cdot 16 - 12 - 1 = 128 - 48 - 12 - 1 = 67 \geq 0$ . The derivative with respect to $k$ is $6k^2 - 6k - 3 = 3(2k^2 - 2k - 1)$ , which is positive for $k \geq 2$ , so $2k^3 - 3k^2 - 3k - 1$ is strictly increasing for $k \geq 4$ . Hence $3k^3 \geq (k+1)^3$ for all $k \geq 4$ .

Step 5. Combine. Chain the inequalities:

3^{k+1} > 3k^3 \geq (k+1)^3.

So $3^{k+1} > (k+1)^3$ and $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $3^n > n^3$ for all integers $n \geq 4$ .

Flavour 4: Recursive sequences with a closed form

A recursively defined sequence $a_1$ , $a_{n+1} = g(a_n)$ is often given with a proposed closed form. Prove the closed form by induction.

Let

a_1 = 1

and

a_{n+1} = 3 a_n + 2

for

n \geq 1

. Prove by induction that

a_n = 2 \cdot 3^{n-1} - 1

for all integers

n \geq 1

Step 1. Base case ( $n = 1$ ): Closed form: $2 \cdot 3^0 - 1 = 2 - 1 = 1$ . Given: $a_1 = 1$ . The values agree, so the statement holds for $n = 1$ .
Step 2. Inductive hypothesis: Assume the closed form holds at $n = k$ , that is, $a_k = 2 \cdot 3^{k-1} - 1$ .
Step 3. Apply the recursion: Use the definition $a_{k+1} = 3 a_k + 2$ and substitute the hypothesis.

a_{k+1} = 3 (2 \cdot 3^{k-1} - 1) + 2 = 6 \cdot 3^{k-1} - 3 + 2 = 2 \cdot 3^{k} - 1.

Step 4. Match the closed form at $n = k + 1$ . The proposed formula gives $a_{k+1} = 2 \cdot 3^{(k+1) - 1} - 1 = 2 \cdot 3^{k} - 1$ . The two expressions agree, so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $a_n = 2 \cdot 3^{n-1} - 1$ for all integers $n \geq 1$ .

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.

Check your knowledge

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

Prove by induction that $\sum_{i=1}^{n} (2i - 1) = n^2$ for all integers $n \geq 1$ .
Prove by induction that $\sum_{i=1}^{n} 2^i = 2^{n+1} - 2$ for all integers $n \geq 1$ .
Prove by induction that $9^n - 1$ is divisible by $8$ for all integers $n \geq 1$ .
Prove by induction that $\sum_{i=1}^{n} i \cdot i! = (n+1)! - 1$ for all integers $n \geq 1$ .
Prove by induction that $11^n - 4^n$ is divisible by $7$ for all integers $n \geq 1$ .
Prove by induction that $2^n \geq n + 1$ for all integers $n \geq 1$ .
Prove by induction that $n! > 2^n$ for all integers $n \geq 4$ .
The sequence is defined by $a_1 = 2$ and $a_{n+1} = 2 a_n + 1$ for $n \geq 1$ . Prove by induction that $a_n = 3 \cdot 2^{n-1} - 1$ for all integers $n \geq 1$ .

Solutions

Q1
Step 1. Base case ( $n = 1$ ): LHS $= 2(1) - 1 = 1$ . RHS $= 1^2 = 1$ . Holds.
Step 2. Inductive hypothesis: Assume $\sum_{i=1}^{k} (2i - 1) = k^2$ .
Step 3. Split off the $(k+1)$ -th term

\sum_{i=1}^{k+1} (2i - 1) = k^2 + (2(k+1) - 1) = k^2 + 2k + 1 = (k+1)^2.

Step 4. Match the target: $(k+1)^2$ is the formula at $n = k + 1$ . So $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $\sum_{i=1}^{n} (2i - 1) = n^2$ for all integers $n \geq 1$ .
Q2
Step 1. Base case ( $n = 1$ ): LHS $= 2^1 = 2$ . RHS $= 2^2 - 2 = 2$ . Holds.
Step 2. Inductive hypothesis: Assume $\sum_{i=1}^{k} 2^i = 2^{k+1} - 2$ .
Step 3. Split off the $(k+1)$ -th term

\sum_{i=1}^{k+1} 2^i = (2^{k+1} - 2) + 2^{k+1} = 2 \cdot 2^{k+1} - 2 = 2^{k+2} - 2.

Step 4. Match the target: The formula at $n = k + 1$ is $2^{(k+1)+1} - 2 = 2^{k+2} - 2$ . So $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $\sum_{i=1}^{n} 2^i = 2^{n+1} - 2$ for all integers $n \geq 1$ .
Q3
Step 1. Base case ( $n = 1$ ): $9^1 - 1 = 8 = 8 \cdot 1$ , divisible by $8$ . Holds.
Step 2. Inductive hypothesis: Assume $9^k - 1 = 8M$ for some integer $M$ . So $9^k = 8M + 1$ .
Step 3. Rewrite at $n = k + 1$

9^{k+1} - 1 = 9 \cdot 9^k - 1 = 9(8M + 1) - 1 = 72M + 9 - 1 = 72M + 8 = 8(9M + 1).

Step 4. Conclude divisibility: Since $9M + 1$ is an integer, $9^{k+1} - 1$ is divisible by $8$ . So $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $9^n - 1$ is divisible by $8$ for all integers $n \geq 1$ .
Q4
Step 1. Base case ( $n = 1$ ): LHS $= 1 \cdot 1! = 1$ . RHS $= 2! - 1 = 1$ . Holds.
Step 2. Inductive hypothesis: Assume $\sum_{i=1}^{k} i \cdot i! = (k+1)! - 1$ .
Step 3. Split off the $(k+1)$ -th term

\sum_{i=1}^{k+1} i \cdot i! = (k+1)! - 1 + (k+1)(k+1)!.

Step 4. Factor $(k+1)!$ .

(k+1)! - 1 + (k+1)(k+1)! = (k+1)! \big[ 1 + (k+1) \big] - 1 = (k+1)! (k+2) - 1 = (k+2)! - 1.

Step 5. Match the target: The formula at $n = k + 1$ is $((k+1) + 1)! - 1 = (k+2)! - 1$ . So $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $\sum_{i=1}^{n} i \cdot i! = (n+1)! - 1$ for all integers $n \geq 1$ .
Q5
Step 1. Base case ( $n = 1$ ): $11^1 - 4^1 = 7 = 7 \cdot 1$ , divisible by $7$ . Holds.
Step 2. Inductive hypothesis: Assume $11^k - 4^k = 7M$ for some integer $M$ . So $11^k = 7M + 4^k$ .
Step 3. Rewrite at $n = k + 1$: Multiply by $11$ and balance.

11^{k+1} - 4^{k+1} = 11 \cdot 11^k - 4 \cdot 4^k = 11(7M + 4^k) - 4 \cdot 4^k = 77M + 11 \cdot 4^k - 4 \cdot 4^k.

Step 4. Combine the $4^k$ terms.

= 77M + 7 \cdot 4^k = 7(11M + 4^k).

Step 5. Conclude divisibility: Since $11M + 4^k$ is an integer, $11^{k+1} - 4^{k+1}$ is divisible by $7$ . So $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $11^n - 4^n$ is divisible by $7$ for all integers $n \geq 1$ .
Q6
Step 1. Base case ( $n = 1$ ): $2^1 = 2$ and $1 + 1 = 2$ , so $2 \geq 2$ . Holds.
Step 2. Inductive hypothesis: Assume $2^k \geq k + 1$ for some $k \geq 1$ .
Step 3. Double the hypothesis: Multiply both sides by $2$ :

2^{k+1} = 2 \cdot 2^k \geq 2(k+1) = 2k + 2.

Step 4. Compare with the target. We want $2^{k+1} \geq (k+1) + 1 = k + 2$ . Since $2k + 2 \geq k + 2$ for all $k \geq 0$ , we have

2^{k+1} \geq 2k + 2 \geq k + 2.

So $P(k) \implies P(k+1)$ .

Conclusion: By the principle of mathematical induction, $2^n \geq n + 1$ for all integers $n \geq 1$ .
Q7
Step 1. Base case ( $n = 4$ ): $4! = 24$ and $2^4 = 16$ . Since $24 > 16$ , the statement holds for $n = 4$ .
Step 2. Inductive hypothesis: Assume $k! > 2^k$ for some $k \geq 4$ .
Step 3. Multiply by $k + 1$: Use the factorial recurrence $(k+1)! = (k+1) \cdot k!$ .

(k+1)! = (k+1) \cdot k! > (k+1) \cdot 2^k.

Step 4. Compare with $2^{k+1}$: Since $k \geq 4$ , we have $k + 1 \geq 5 > 2$ , so $(k+1) \cdot 2^k > 2 \cdot 2^k = 2^{k+1}$ .
Step 5. Combine: Therefore $(k+1)! > 2^{k+1}$ , so $P(k) \implies P(k+1)$ .
Conclusion: By the principle of mathematical induction, $n! > 2^n$ for all integers $n \geq 4$ .
Q8
Step 1. Base case ( $n = 1$ ): Closed form: $3 \cdot 2^{0} - 1 = 3 - 1 = 2$ . Given: $a_1 = 2$ . Agrees.
Step 2. Inductive hypothesis: Assume $a_k = 3 \cdot 2^{k-1} - 1$ for some $k \geq 1$ .
Step 3. Apply the recursion

a_{k+1} = 2 a_k + 1 = 2 (3 \cdot 2^{k-1} - 1) + 1 = 3 \cdot 2^{k} - 2 + 1 = 3 \cdot 2^{k} - 1.

Step 4. Match the closed form at $n = k + 1$ . The proposed formula gives $a_{k+1} = 3 \cdot 2^{(k+1)-1} - 1 = 3 \cdot 2^{k} - 1$ . Agrees, so $P(k) \implies P(k+1)$ .

Conclusion. By the principle of mathematical induction, $a_n = 3 \cdot 2^{n-1} - 1$ for all integers $n \geq 1$ .

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.

Sources & how we know this

Mathematics Extension 1 Stage 6 Syllabus (2017) — NESA
Mathematics Extension 1 HSC exam papers — NESA