Worked Solutions
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.
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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.