QCE Vault / Specialist Maths Mathematical induction and trigonometric proofs — Question 180 Original QCE Vault practice · 6 marks
Browse all questions Original practice exam Report an issue Q180 · Practice question Technology-free Complex familiar 6 marks
QUESTION 180 (6 marks) For positive integers n, consider 8 n − 1 8^n-1 8 n − 1 . a) Prove by induction that this number is divisible by 7.
[4 marks] b) Show ( 8 n − 1 ) / 7 = 1 + 8 + ⋯ + 8 n − 1 (8^n-1)/7=1+8+\cdots+8^{n-1} ( 8 n − 1 ) /7 = 1 + 8 + ⋯ + 8 n − 1 . [2 marks] WORKED SOLUTION
Practice marking scheme 6 marks ANSWER (a) 7 ∣ 8 n − 1 7\mid8^n-1 7 ∣ 8 n − 1 . (b) Geometric-sum identity established. Worked solution
(a) At n=1 the value is 7. Assume 8 m − 1 = 7 r 8^m-1=7r 8 m − 1 = 7 r for an integer r. Then 8 m + 1 − 1 = 8 ( 8 m − 1 ) + 7 = 7 ( 8 r + 1 ) 8^{m+1}-1=8(8^m-1)+7=7(8r+1) 8 m + 1 − 1 = 8 ( 8 m − 1 ) + 7 = 7 ( 8 r + 1 ) , also divisible by 7. The induction principle proves the claim for all positive integers. (b) Let G = 1 + 8 + ⋯ + 8 n − 1 G=1+8+\cdots+8^{n-1} G = 1 + 8 + ⋯ + 8 n − 1 . Subtracting G from 8G cancels all intermediate terms and gives 7 G = 8 n − 1 7G=8^n-1 7 G = 8 n − 1 . Divide by 7. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: State an integer induction assumption.
[1 mark] Part a: Express the next case as 7 times an integer.
[1 mark] Part a: Conclude for all positive integers.
[1 mark] Part b: Multiply the sum by 8 and subtract.
[1 mark] Part b: Obtain the quotient identity.
[1 mark] Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.
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