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Specialist Mathematics · Past QCAA questions All 24 questions
2022 · Paper 1 · Technology-free

Question 18

[5 marks] Technology-free
Unit 3 · Topic 2Mathematical induction and trigonometric proofs Tool Kit 2.2Divisibility proofs Tool Kit 2.1Summation proofs (step 3)
The question

It is proposed that the following expression is divisible by $\left(1+\operatorname{cis}(\theta)\right)$ for $n\in\mathbb{Z}^{+}$, $\left(1+\operatorname{cis}(\theta)\right)\neq0$.

$$\sum_{r=0}^{2n+1}\operatorname{cis}(r\theta)$$

Evaluate the reasonableness of the proposition.

Watch the proof first

Follow each part of the sum from line to line

Watch the five moves of an induction proof, then the inductive step one line at a time. Blue is the $n=k$ sum and orange is the two new terms.

Every induction proof uses the same five moves Moves 1 and 2 done. Start with the n = k + 1 sum Split it into the n = k sum and two new terms Your assumption replaces the n = k sum Take (cis θ)²ᵏ⁺² out of the two new terms Now 1 + cis θ is a common factor True for n = k + 1, so true for all n ∈ ℤ⁺ 1 Show n = 1 is true mark 1 2 Assume n = k is true mark 2 3 Write n = k + 1 using the assumption mark 3 4 Find the common factor mark 4 ★ 5 Conclude by induction mark 5 ✓ n = 1 is true ✓ assume Σ cis(rθ) to 2k + 1 = (1 + cis θ)Q(θ) Σ 2k + 3 r = 0 cis(rθ) n = k + 1 = Σ 2k + 1 r = 0 cis(rθ) + cis((2k + 2)θ) + cis((2k + 3)θ) the n = k sum the two new terms mark 3 = (1 + cis θ)Q(θ) + (cis θ)2k+2 + (cis θ)2k+3 De Moivre: cis(mθ) = (cis θ)ᵐ mark 3 = (1 + cis θ)Q(θ) + (cis θ)2k+2(1 + cis θ) mark 4 ★ = (1 + cis θ)(Q(θ) + (cis θ)2k+2) call the second bracket R(θ) mark 5 so 1 + cis θ divides the sum for every n ∈ ℤ⁺

The orange line is the insight mark. Once the new terms show $1+\operatorname{cis}(\theta)$, the rest of the proof follows.

Now for the mathematics

Build the inductive step yourself

Answer each question before you look at the marking guide. These are the decisions you have to make in the exam.

5. Put the five moves of the proof in order.
QCAA marking guide · 5 marks

Work through the proof one mark at a time

Try each step yourself before you reveal it. You use mathematical induction, and you need De Moivre’s theorem, $\operatorname{cis}(r\theta)=\left(\operatorname{cis}(\theta)\right)^{r}$, all the way through.

Step 11 mark

Start with $n=1$. The upper limit is $2(1)+1=3$, so you add four terms. Write each one as a power of $\operatorname{cis}(\theta)$.

$$\begin{aligned}\sum_{r=0}^{3}\operatorname{cis}(r\theta)&=1+\operatorname{cis}(\theta)+\left(\operatorname{cis}(\theta)\right)^{2}+\left(\operatorname{cis}(\theta)\right)^{3}\\&=\left(1+\operatorname{cis}(\theta)\right)+\left(\operatorname{cis}(\theta)\right)^{2}\left(1+\operatorname{cis}(\theta)\right)\\&=\left(1+\operatorname{cis}(\theta)\right)\left(1+\left(\operatorname{cis}(\theta)\right)^{2}\right)\end{aligned}$$

This has a factor of $1+\operatorname{cis}(\theta)$, so the proposition is true for $n=1$. Remember that $\operatorname{cis}(0)=1$.

Marker: correctly proves the initial statement

Step 21 mark

Assume the proposition is true for $n=k$, where $k\in\mathbb{Z}^{+}$. Write the assumption as an equation, with $Q(\theta)$ standing for whatever is left after taking out the factor.

$$\sum_{r=0}^{2k+1}\operatorname{cis}(r\theta)=\left(1+\operatorname{cis}(\theta)\right)Q(\theta)$$

You can also say in words that the sum has a factor of $1+\operatorname{cis}(\theta)$, but the equation is much easier to use in the next step.

Marker: correctly establishes an appropriate assumption for $n=k$

Step 31 mark

Now let $n=k+1$. The upper limit becomes $2(k+1)+1=2k+3$, so you get the $n=k$ sum plus two extra terms.

$$\begin{aligned}\sum_{r=0}^{2k+3}\operatorname{cis}(r\theta)&=\sum_{r=0}^{2k+1}\operatorname{cis}(r\theta)+\operatorname{cis}\left((2k+2)\theta\right)+\operatorname{cis}\left((2k+3)\theta\right)\\&=\left(1+\operatorname{cis}(\theta)\right)Q(\theta)+\left(\operatorname{cis}(\theta)\right)^{2k+2}+\left(\operatorname{cis}(\theta)\right)^{2k+3}\end{aligned}$$

You replaced the first part with your assumption. You keep follow-through marks from here if you slipped earlier.

Marker: expresses the sum based on $n=k+1$ in terms of the assumption

Step 4 · insight mark ★1 mark

Take out $\left(\operatorname{cis}(\theta)\right)^{2k+2}$ from the two new terms. What is left is exactly the factor you need.

$$\left(\operatorname{cis}(\theta)\right)^{2k+2}+\left(\operatorname{cis}(\theta)\right)^{2k+3}=\left(\operatorname{cis}(\theta)\right)^{2k+2}\left(1+\operatorname{cis}(\theta)\right)$$

$$\sum_{r=0}^{2k+3}\operatorname{cis}(r\theta)=\left(1+\operatorname{cis}(\theta)\right)Q(\theta)+\left(\operatorname{cis}(\theta)\right)^{2k+2}\left(1+\operatorname{cis}(\theta)\right)$$

This is the orange line in the animation. Once the two new terms show $1+\operatorname{cis}(\theta)$, you have a common factor.

Marker: expresses a result using a common factor of $\left(1+\operatorname{cis}(\theta)\right)$

Step 51 mark

Take out the common factor, and give the second bracket a name.

$$\begin{aligned}\sum_{r=0}^{2k+3}\operatorname{cis}(r\theta)&=\left(1+\operatorname{cis}(\theta)\right)\left(Q(\theta)+\left(\operatorname{cis}(\theta)\right)^{2k+2}\right)\\&=\left(1+\operatorname{cis}(\theta)\right)R(\theta)\end{aligned}$$

So the proposition is true for $n=k+1$ whenever it is true for $n=k$.

The proposition is reasonable. By mathematical induction, $1+\operatorname{cis}(\theta)$ divides the sum for all $n\in\mathbb{Z}^{+}$.

The question asks you to evaluate reasonableness, so finish with a sentence that says the proposition is true and that you proved it by induction.

Marker: proves the inductive step

Putting it all together

Each phrase of the question gave you something

“divisible by $\left(1+\operatorname{cis}(\theta)\right)$”
You need to write the sum as $1+\operatorname{cis}(\theta)$ times another expression.
“for $n\in\mathbb{Z}^{+}$”
You use induction, starting at $n=1$.
“$\sum_{r=0}^{2n+1}$”
The sum has $2n+2$ terms, so each step up in $n$ adds two terms.
“$\operatorname{cis}(r\theta)$”
You rewrite each term as $\left(\operatorname{cis}(\theta)\right)^{r}$ with De Moivre’s theorem, so you can factorise.
“Evaluate the reasonableness”
You finish by saying clearly that the proposition is true.
What makes this complex unfamiliar

Most divisibility proofs you have seen use whole numbers. Here the thing you divide by is a complex expression, and the common factor only appears once you rewrite $\operatorname{cis}(r\theta)$ as a power of $\operatorname{cis}(\theta)$.

The line I would check first is where you move from $n=k$ to $n=k+1$. The upper limit jumps from $2k+1$ to $2k+3$, so you add two terms. The QCAA like sums where each step adds more than one term, so you need to write out the $n=k+1$ case carefully.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2022 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2022, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.