Principle of Mathematical Induction

(k+1)/(k+2)

Step 1: Take LCM of (k+1) and (k+1)(k+2), which is (k+1)(k+2). Step 2: k/(k+1) = k(k+2)/[(k+1)(k+2)]. Step 3: Adding: [k(k+2) + 1] / [(k+1)(k+2)] = [k² + 2k + 1] / [(k+1)(k+2)]. Step 4: k² + 2k + 1 = (k+1)², so the expression = (k+1)² / [(k+1)(k+2)] = (k+1)/(k+2). Step 5: This equals the RHS of P(k+1), completing the induction step. Recognising (k+1)² in the numerator is key.

2k ≥ k + 1 for k ≥ 1

Step 1: Assume P(k): 2ᵏ > k is true. Step 2: Multiply both sides by 2: 2^(k+1) > 2k. Step 3: We need to show 2^(k+1) > k + 1. Since 2k = k + k and k ≥ 1, we have 2k ≥ k + 1. Step 4: Therefore 2^(k+1) > 2k ≥ k + 1, which gives 2^(k+1) > k + 1. Step 5: This proves P(k+1) is true. The key step is recognising that 2k ≥ k+1 for all natural numbers k ≥ 1.

P(n) is true for all n if P(1) is true AND P(k) true implies P(k+1) true

Step 1: The Principle of Mathematical Induction has two essential conditions. Step 2: Condition 1 (Base Step): P(1) must be true. Step 3: Condition 2 (Induction Step): Assuming P(k) is true, we must prove P(k+1) is true. Step 4: BOTH conditions are necessary. If only P(1) is true but the induction step fails, the proof is invalid. If only the induction step holds but P(1) is false, the proof is invalid. Step 5: Option 2 is wrong because the implication goes forward (k to k+1), not backward. Option 4 is missing the base case.

6, which is divisible by 6

Step 1: Substitute n = 1 in the expression n³ + 5n. Step 2: 1³ + 5(1) = 1 + 5 = 6. Step 3: 6 ÷ 6 = 1, so 6 is exactly divisible by 6. Step 4: Therefore P(1) is true, confirming the base case. Step 5: A common error is computing 1³ + 5 = 6 correctly but then doubting whether 6 is divisible by 6 — remember divisibility means the remainder is 0.