Proof — Worked Solutions (HSC Maths Extension 1)
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Worked examples for HSC Maths Extension 1 proof by mathematical induction. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for HSC Maths Extension 1 — Proof. Every question is worked through with the method and reasoning shown, in NESA exam style, so you can check how to reach the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Induction for a series sum
Question
Prove by mathematical induction that for all integers $n \geq 1$, $$1 \cdot 2 + 2 \cdot 3 + 3 \cdot 4 + \cdots + n(n+1) = \frac{n(n+1)(n+2)}{3}.$$
Solution
Base case $n = 1$: LHS $= 1 \cdot 2 = 2$; RHS $= \dfrac{1 \cdot 2 \cdot 3}{3} = 2$. True.
Assume true for $n = k$: $\sum_{r=1}^{k} r(r+1) = \dfrac{k(k+1)(k+2)}{3}$.
Prove for $n = k+1$. Add the next term $(k+1)(k+2)$:
$$\frac{k(k+1)(k+2)}{3} + (k+1)(k+2) = (k+1)(k+2)\!\left[\frac{k}{3} + 1\right] = \frac{(k+1)(k+2)(k+3)}{3}.$$
That is the formula with $n = k+1$. By induction it holds for all $n \geq 1$. Always state the conclusion sentence — markers want it.
Induction has three steps, and naming them keeps you organised.
Step 1 — base case. Check $n = 1$: the left side is $1 \cdot 2 = 2$, and the right side is $\dfrac{1(2)(3)}{3} = 2$. They match, so the statement is true to start with.
Step 2 — assumption. Suppose it works for some integer $n = k$, i.e. $\sum_{r=1}^{k} r(r+1) = \dfrac{k(k+1)(k+2)}{3}$. This is our stepping stone.
Step 3 — inductive step. For $n = k+1$ we just add the next term, $(k+1)(k+2)$, to the assumed sum:
$$\frac{k(k+1)(k+2)}{3} + (k+1)(k+2).$$
The trick is to factor out the common $(k+1)(k+2)$ rather than expand everything:
$$(k+1)(k+2)\left[\frac{k}{3} + 1\right] = (k+1)(k+2)\cdot\frac{k+3}{3} = \frac{(k+1)(k+2)(k+3)}{3}.$$
That is exactly the formula with $k+1$ in place of $n$. So if it is true for $k$ it is true for $k+1$, and since it is true for $n = 1$, it is true for all $n \geq 1$. The factoring works because it lets the pattern reveal itself instead of drowning in algebra.
- Base $n=1$: LHS $=2$, RHS $=\frac{1\cdot2\cdot3}{3}=2$ ✓
- Assume $n=k$: $\sum_{r=1}^{k} r(r+1)=\frac{k(k+1)(k+2)}{3}$
- Step $n=k+1$: add $(k+1)(k+2)$: $$\frac{k(k+1)(k+2)}{3}+(k+1)(k+2)=(k+1)(k+2)\cdot\frac{k+3}{3}=\frac{(k+1)(k+2)(k+3)}{3}$$
- Matches formula for $k+1$ ⇒ true $\forall n\geq1$.
Where the marks go
- 1 mark: Verifies the base case $n = 1$
- 1 mark: States the inductive assumption and adds the $(k+1)$ term
- 1 mark: Factorises to reach $\frac{(k+1)(k+2)(k+3)}{3}$ and concludes
Key idea
Prove the base case, assume the result for $n = k$, then add the next term and factor out the common product to reach the $n = k+1$ form.
Example 2 — Induction for divisibility
Question
Prove by mathematical induction that $7^n - 3^n$ is divisible by $4$ for all integers $n \geq 1$.
Solution
Base case $n = 1$: $7 - 3 = 4$, divisible by 4. True.
Assume $7^k - 3^k = 4M$ for some integer $M$.
Prove $7^{k+1} - 3^{k+1}$ is divisible by 4. Write $7^{k+1} = 7 \cdot 7^k$ and split:
$$7^{k+1} - 3^{k+1} = 7\cdot 7^k - 3\cdot 3^k = 7(7^k - 3^k) + 7\cdot 3^k - 3\cdot 3^k.$$
So $7^{k+1} - 3^{k+1} = 7(4M) + 4\cdot 3^k = 4(7M + 3^k)$.
That is $4 \times$ an integer, so divisible by 4. By induction it holds for all $n \geq 1$. The move that earns marks is rewriting $7^{k+1}$ so the assumed term $7^k - 3^k$ appears.
Divisibility induction looks scary but follows the same three steps.
Base case. For $n = 1$, $7^1 - 3^1 = 4$, which is clearly divisible by 4.
Assumption. Suppose for $n = k$ that $7^k - 3^k = 4M$, where $M$ is some integer. We do not need to know $M$ — just that it exists.
Inductive step. We want to show $7^{k+1} - 3^{k+1}$ is also a multiple of 4. The key is to make the assumed expression appear. Write $7^{k+1} = 7 \cdot 7^k$ and add-and-subtract so $7^k - 3^k$ surfaces:
$$7^{k+1} - 3^{k+1} = 7\cdot7^k - 3\cdot3^k = 7\underbrace{(7^k - 3^k)}_{=4M} + (7-3)3^k.$$
Now substitute the assumption: $= 7(4M) + 4\cdot3^k = 4(7M + 3^k)$.
Since $7M + 3^k$ is an integer, the whole thing is a multiple of 4. So truth at $k$ forces truth at $k+1$; with the base case done, it holds for all $n \geq 1$. The add-and-subtract trick works because it lets us reuse what we already know is divisible by 4.
- Base $n=1$: $7-3=4$ ✓
- Assume $7^k-3^k=4M$, $M\in\mathbb{Z}$
- Step: $$7^{k+1}-3^{k+1}=7\cdot7^k-3\cdot3^k=7(7^k-3^k)+4\cdot3^k$$ $$=7(4M)+4\cdot3^k=4(7M+3^k)$$
- Multiple of 4 ⇒ true $\forall n\geq1$.
Where the marks go
- 1 mark: Verifies the base case $n = 1$ gives $4$
- 1 mark: States the assumption $7^k - 3^k = 4M$
- 1 mark: Rewrites $7^{k+1} - 3^{k+1}$ to expose the assumed term
- 1 mark: Factors out 4 and concludes by induction
Key idea
For divisibility induction, rewrite the $k+1$ expression so the assumed multiple appears, then factor out the divisor.
Frequently asked questions
Step-by-step solutions to exam-style questions on Proof in HSC Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions in the exam, rather than memorising answers.
Yes — they follow the NESA HSC Maths Extension 1 syllabus for Proof, using the methods and notation expected in the exam.