Mathematical induction and trigonometric proofs — Question 179
Original QCE Vault practice · 6 marks
Q179 · Practice questionTechnology-freeComplex familiar6 marks
QUESTION 179 (6 marks)
For positive integers n, let Sn=∑k=1n(2k−1)3. a)Prove by induction that Sn=n2(2n2−1). b)Determine the least n for which the sum exceeds 1000, with justification.
WORKED SOLUTION
Practice marking scheme
6 marksANSWER(a) Identity proved. (b) n=5. Worked solution
(a) For n=1 both sides are 1. Assume Sm=m2(2m2−1). Then Sm+1=m2(2m2−1)+(2m+1)3=2m4+8m3+11m2+6m+1=(m+1)2(2(m+1)2−1). This establishes the next case, hence the result for every positive integer by induction.
(b) S4=496 and S5=1225. Every added odd cube is positive, so the sum strictly increases and 5 is the least qualifying n. Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: Verify the base case.
Part a: State the induction assumption.
Part a: Derive the next case algebraically.
Part a: Conclude by induction.
Part b: Evaluate the adjacent sums.
Part b: Use monotonicity to establish minimality.
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