QCE Vault / Specialist Maths Proof by mathematical induction — Question 3 QCAA 2024, Paper 1 · 1 mark
Browse all questions Original practice exam Report an issue Q3 · 2024 · Technology-free Simple familiar 1 mark
QUESTION 3 Consider a proof of the proposition ∑ j = 1 n ( 2 j − 1 ) = n 2 ∀ n ∈ Z + \displaystyle\sum_{j=1}^{n}(2j-1)=n^2\ \forall n\in\mathbb Z^+ j = 1 ∑ n ( 2 j − 1 ) = n 2 ∀ n ∈ Z + using mathematical induction. Within the proof of the inductive step, the proposition for n = k + 1 n=k+1 n = k + 1 could be expressed as (A) ∑ j = 1 k + 1 ( 2 j − 1 ) = k 2 + 2 k + 1 \displaystyle\sum_{j=1}^{k+1}(2j-1)=k^2+2k+1 j = 1 ∑ k + 1 ( 2 j − 1 ) = k 2 + 2 k + 1 (B) ∑ j = 1 k + 1 ( 2 k + 1 ) = k 2 + 2 k + 1 \displaystyle\sum_{j=1}^{k+1}(2k+1)=k^2+2k+1 j = 1 ∑ k + 1 ( 2 k + 1 ) = k 2 + 2 k + 1 (C) ∑ j = 1 k + 1 ( 2 j − 1 ) = k 2 + 1 \displaystyle\sum_{j=1}^{k+1}(2j-1)=k^2+1 j = 1 ∑ k + 1 ( 2 j − 1 ) = k 2 + 1 (D) ∑ j = 1 k + 1 ( 2 k + 1 ) = k 2 + 1 \displaystyle\sum_{j=1}^{k+1}(2k+1)=k^2+1 j = 1 ∑ k + 1 ( 2 k + 1 ) = k 2 + 1 WORKED SOLUTION
Answer A 1 mark Worked solution
For n = k + 1 n=k+1 n = k + 1 , the left side becomes ∑ j = 1 k + 1 ( 2 j − 1 ) \sum_{j=1}^{k+1}(2j-1) ∑ j = 1 k + 1 ( 2 j − 1 ) and the right side becomes ( k + 1 ) 2 = k 2 + 2 k + 1 (k+1)^2=k^2+2k+1 ( k + 1 ) 2 = k 2 + 2 k + 1 . Worked explanation by QCE Vault; correct option verified against the QCAA answer key.
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