QCE Vault / Specialist Maths Proof by mathematical induction — Question 4 QCAA 2022, Paper 1 · 1 mark
Browse all questions Original practice exam Report an issue Q4 · 2022 · Technology-free Simple familiar 1 mark
QUESTION 4 When using proof by mathematical induction to prove De Moivre’s theorem expressed as ( r cis ( θ ) ) n = r n cis ( n θ ) (r\operatorname{cis}(\theta))^n=r^n\operatorname{cis}(n\theta) ( r cis ( θ ) ) n = r n cis ( n θ ) ∀ n ∈ Z + \forall n\in\mathbb Z^+ ∀ n ∈ Z + , which statement would be correct in the proof of the inductive step? (A) ( r cis ( θ ) ) k = r k cis ( k θ ) (r\operatorname{cis}(\theta))^k=r^k\operatorname{cis}(k\theta) ( r cis ( θ ) ) k = r k cis ( k θ ) (B) ( r cis ( θ ) ) k = r k + 1 cis ( k + θ ) (r\operatorname{cis}(\theta))^k=r^{k+1}\operatorname{cis}(k+\theta) ( r cis ( θ ) ) k = r k + 1 cis ( k + θ ) (C) ( r cis ( θ ) ) k + 1 = r k + 1 cis ( k θ + 1 ) (r\operatorname{cis}(\theta))^{k+1}=r^{k+1}\operatorname{cis}(k\theta+1) ( r cis ( θ ) ) k + 1 = r k + 1 cis ( k θ + 1 ) (D) ( r cis ( θ ) ) k + 1 = r k + 1 cis ( ( k + 1 ) θ ) (r\operatorname{cis}(\theta))^{k+1}=r^{k+1}\operatorname{cis}((k+1)\theta) ( r cis ( θ ) ) k + 1 = r k + 1 cis (( k + 1 ) θ ) WORKED SOLUTION
Answer D 1 mark Worked solution
The inductive step aims to establish the proposition for n = k + 1 n=k+1 n = k + 1 . Substituting k + 1 k+1 k + 1 into De Moivre’s statement gives ( r cis θ ) k + 1 = r k + 1 cis ( ( k + 1 ) θ ) (r\operatorname{cis}\theta)^{k+1}=r^{k+1}\operatorname{cis}((k+1)\theta) ( r cis θ ) k + 1 = r k + 1 cis (( k + 1 ) θ ) . Worked explanation by QCE Vault; correct option verified against the QCAA answer key.
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