Proof by mathematical induction — Question 18

QCAA 2022, Paper 1 · 5 marks

Q18 · 2022 · Technology-freeComplex unfamiliar5 marks

QUESTION 18 (5 marks)

It is proposed that the following expression is divisible by

1+\operatorname{cis}(\theta)

for

n\in\mathbb Z^+

1+\operatorname{cis}(\theta)\ne0

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

Evaluate the reasonableness of the proposition.

WORKED SOLUTION

5 marks

ANSWER

The proposition is true for all

n\in\mathbb Z^+

Worked solution

Mathematical induction can be used to prove the proposition. For

n=1

\sum_{r=0}^{3}\operatorname{cis}(r\theta) =1+\operatorname{cis}(\theta)+\operatorname{cis}^2(\theta)+\operatorname{cis}^3(\theta) =(1+\operatorname{cis}(\theta))(1+\operatorname{cis}^2(\theta)).

Assume for

n=k

that

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

Then for

n=k+1

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

which has the common factor

1+\operatorname{cis}(\theta)

. Thus the inductive step holds, so the proposition is true by mathematical induction.

correctly proves the initial statement

[1 mark]

correctly establishes an appropriate assumption for

n=k

[1 mark]

expresses the sum based on

n=k+1

in terms of the assumption

[1 mark]

expresses a result using a common factor of

1+\operatorname{cis}(\theta)

[1 mark]

proves the inductive step

[1 mark]

One QCAA sample method typeset for web; criterion wording is adapted from the official marking guide.

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