QCE Specialist Mathematics - Unit 3 - Mathematical induction and trigonometric proofs

Mathematical Induction | QCE Specialist Mathematics

Learn Specialist Mathematics induction proofs for sums, divisibility and De Moivre's theorem, with the exact proof structure QCAA expects.

Updated 2026-05-18 - 5 min read

Mathematical induction is a proof technique for statements indexed by positive integers. It is a formal way of saying: prove the first case, then prove that each true case forces the next one.

The proof structure

A complete induction proof has four pieces:

1. Define the proposition $P(n)$.
2. Prove the initial statement, usually $P(1)$.
3. Assume $P(k)$ is true for some positive integer $k$.
4. Use that assumption to prove $P(k+1)$, then conclude.

The base case must match the first value in the statement. If the result says $n\ge1$, prove $P(1)$. If it says $n\ge0$, prove $P(0)$. Do not automatically start at $1$ just because many examples do.

The inductive step should clearly state what you are trying to prove. A clean layout is:

$ \text{Assume }P(k)\text{ is true for some }k\ge1. $

Then write the $P(k+1)$ statement underneath. This makes it obvious that you know the target, not just the assumption.

Sums

For a sum proof, the trick is to isolate the old sum inside the new one.

Suppose you want to prove:

$ 1+2+\cdots+n=\frac{n(n+1)}{2}. $

Assume:

$ 1+2+\cdots+k=\frac{k(k+1)}{2}. $

Then:

$ 1+2+\cdots+k+(k+1) =\frac{k(k+1)}{2}+(k+1). $

Factor the new expression:

$ =\frac{(k+1)(k+2)}{2}. $

That is the formula with $n=k+1$.

Sigma notation is just a compact way of writing the same idea:

$ \sum_{r=1}^{n} r^2=1^2+2^2+\cdots+n^2. $

So the step from $k$ to $k+1$ is:

$ \sum_{r=1}^{k+1} r^2=\sum_{r=1}^{k} r^2+(k+1)^2. $

That one line is often the moment where the induction assumption becomes useful.

Divisibility

For divisibility, write the assumption as a multiple. If you assume $P(k)$ says $A_k$ is divisible by $m$, then write:

$ A_k=mq $

for some integer $q$. The goal is to rearrange $A_{k+1}$ until it becomes $m$ times another integer.

Write divisibility statements in a form that is easy to manipulate. "Divisible by $7

De Moivre by induction

The same structure proves De Moivre's theorem:

$ [\operatorname{cis}(\theta)]^n=\operatorname{cis}(n\theta). $

The inductive step multiplies by another $\operatorname{cis}(\theta)$ and uses the angle-addition identity.

This is a good example of induction being more than a calculation trick. The base case proves the formula starts correctly, and the inductive step proves that multiplying by one more copy of $\operatorname{cis}(\theta)$ advances the angle by another $\theta$. Once those two pieces are locked together, every positive integer power follows.

A fuller sum example

A classic sum result is:

$ 1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}. $

The base case is $n=1$, where both sides equal $1$. For the inductive step, assume:

$ 1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}. $

Then the next sum is:

$ 1^2+2^2+\cdots+k^2+(k+1)^2. $

Substitute the assumption:

$ \frac{k(k+1)(2k+1)}{6}+(k+1)^2. $

Factor $k+1$ and put everything over $6$:

$ \frac{(k+1)[k(2k+1)+6(k+1)]}{6}. $

The bracket simplifies to:

$ 2k^2+7k+6=(k+2)(2k+3), $

so the expression becomes:

$ \frac{(k+1)(k+2)(2k+3)}{6}. $

That is exactly the formula with $n=k+1$.

Divisibility strategy

Divisibility proofs often feel less predictable than sum proofs because you may need to rearrange the assumption. If the assumption says $A_k$ is divisible by $m$, write:

$ A_k=mq $

for some integer $q$, then solve for the awkward expression you expect to need. After expanding $A_{k+1}$, look for a piece that resembles $A_k$. Substitute the assumption once, then factor out $m$.

If you use the assumption multiple times, pause and check the algebra. Most QCE divisibility proofs are designed so the assumption unlocks the expression once, then the rest is integer arithmetic.

Trigonometric induction

Some induction proofs use trigonometric identities rather than polynomial algebra. For example, a common style is to prove:

$ \sum_{r=1}^{n}\cos(r\theta) = \frac{\sin\left(\frac{n\theta}{2}\right)\cos\left(\frac{(n+1)\theta}{2}\right)} {\sin\left(\frac{\theta}{2}\right)} $

for values of $\theta$ where $\sin\left(\frac{\theta}{2}\right)\ne0$.

The inductive step starts from the $k+1$ sum:

$ \sum_{r=1}^{k+1}\cos(r\theta) =\sum_{r=1}^{k}\cos(r\theta)+\cos((k+1)\theta). $

Substitute the assumption for the old sum, then combine terms using product-to-sum identities. The key identity is:

$ 2\sin A\cos B=\sin(A+B)+\sin(A-B). $

This kind of proof feels different from a sum of powers, but the logic is identical. The new left-hand side contains the old left-hand side plus one extra trig term. The only extra work is choosing the identity that reshapes the expression into the $k+1$ target.

De Moivre's theorem also fits this structure. Assuming:

$ \operatorname{cis}^k\theta=\operatorname{cis}(k\theta), $

the next case is:

$ \operatorname{cis}^{k+1}\theta =\operatorname{cis}^k\theta\operatorname{cis}\theta =\operatorname{cis}(k\theta)\operatorname{cis}\theta =\operatorname{cis}((k+1)\theta). $

The final equality uses angle addition, not guesswork.

Harder divisibility moves

Harder divisibility proofs often need you to rewrite the $k+1$ expression so the $k$ assumption appears inside it. Suppose you want to prove $7\mid(8^n-1)$. The assumption is:

$ 8^k-1=7q $

for some integer $q$. For the next case:

$ 8^{k+1}-1=8\cdot8^k-1. $

Now insert and subtract $8$:

$ 8\cdot8^k-1=8(8^k-1)+7. $

The assumption applies to $8^k-1$, so:

$ 8(8^k-1)+7=8(7q)+7=7(8q+1). $

That is divisible by $7$. The useful move was not expanding harder; it was forcing the expression to contain the old case.

Quick check

Sources

• QCAA Specialist Mathematics subject page
• QCAA Specialist Mathematics 2025 syllabus
• LibreTexts: Mathematical Induction
• LibreTexts: Divisibility Statements and Proofs Using Induction