Proof by Mathematical Induction (Extension) — Topic Review
This review covers all three lessons in the Proof by Mathematical Induction (Extension) topic: Induction for Series and Sums, Induction for Divisibility and Inequalities, and Induction for Matrices and Complex Numbers. Questions are exam-style and increase in difficulty.
Mixed Review Questions
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Fluency
Q1 — Sum of Integers
Prove by mathematical induction that ∑k=1n k = n(n+1)/2 for all positive integers n.
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Fluency
Q2 — Base Case for Divisibility
Verify the base case for the statement: “3n − 1 is divisible by 2 for all n ≥ 1.”
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Fluency
Q3 — Inductive Hypothesis for Sum of Squares
Write out the inductive hypothesis for proving ∑k=1n k² = n(n+1)(2n+1)/6 by mathematical induction.
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Fluency
Q4 — De Moivre’s Theorem Application
Evaluate (cos(π/4) + i sin(π/4))8 using De Moivre’s theorem.
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Fluency
Q5 — Sum of Odd Integers
Prove by induction that ∑k=1n (2k − 1) = n² for all positive integers n.
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Understanding
Q6 — Sum of Squares
Prove by mathematical induction that ∑k=1n k² = n(n+1)(2n+1)/6 for all positive integers n.
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Understanding
Q7 — Divisibility by 3
Prove by induction that 4n − 1 is divisible by 3 for all n ≥ 1.
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Understanding
Q8 — Factorial Inequality
Prove by induction that n! > 2n for all n ≥ 4.
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Understanding
Q9 — Power of a Matrix
Prove by induction that ⎛1 2⎞n = ⎛1 2n⎞
⎝0 1⎠ ⎝0 1⎠
for all positive integers n. -
Understanding
Q10 — De Moivre’s Theorem (Proof)
Prove De Moivre’s theorem for positive integers: (cosθ + i sinθ)n = cos(nθ) + i sin(nθ) for all n ∈ ℤ+.
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Problem Solving
Q11 — Power vs Square Inequality
Prove by induction that 2n > n² for all n ≥ 5.
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Problem Solving
Q12 — Diagonal Matrix Power
Prove by induction that ⎛3 0⎞n = ⎛3n 0 ⎞
⎝0 2⎠ ⎝0 2n⎠
for all positive integers n. -
Problem Solving
Q13 — Triple Angle Formula
Use De Moivre’s theorem with n = 3 to prove that cos(3θ) = 4cos³θ − 3cosθ.
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Problem Solving
Q14 — Sum Involving Powers of 2
Prove by induction that ∑k=1n k · 2k = (n − 1) · 2n+1 + 2 for all positive integers n.
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Problem Solving
Q15 — Telescoping Series
Prove by induction that 1/(1·2) + 1/(2·3) + … + 1/(n(n+1)) = n/(n+1) for all positive integers n.