Principle of Mathematical Induction

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NOT a statement. Reason: A statement must be either true or false, but not both. Since we don't know the value of x, we cannot determine if this is true or false. It becomes a statement only if we spe…

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YES, this is a statement. Reason: Even though n is a variable, this is a universal statement meaning 'for all natural numbers n, 1 + 2 + 3 + ... + n = n(n+1)/2'. It is either true or false (in this ca…

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P(1): 2^1 > 1 - 1, which is 2 > 0 (TRUE) P(k): 2^k > k - 1 P(k+1): 2^(k+1) > (k+1) - 1, which is 2^(k+1) > k Method: Replace n with 1, k, and k+1 in the original statement. This step is crucial for s…

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P(1): 1 = 1(3·1-1)/2 = 1·2/2 = 1 ✓ P(2): 1 + 4 = 2(3·2-1)/2 = 2·5/2 = 5 ✓ P(k+1): 1 + 4 + 7 + ... + (3k-2) + [3(k+1)-2] = (k+1)[3(k+1)-1]/2 Simplified: 1 + 4 + 7 + ... + (3k+1) = (k+1)(3k+2)/2 Key: T…

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P(1): 6 divides (1³ + 5·1) = 6 divides 6 ✓ (TRUE) P(k): 6 divides (k³ + 5k) P(k+1): 6 divides [(k+1)³ + 5(k+1)] Expanded P(k+1): 6 divides [k³ + 3k² + 3k + 1 + 5k + 5] = 6 divides [k³ + 3k² + 8k + 6]…

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Condition 1 (Base Case): P(1) must be TRUE - Verify that the statement is true when n = 1 - This establishes the starting point of the proof Condition 2 (Inductive Step): IF P(k) is true, THEN P(k+1)…

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STEP 1 (BASE CASE): Show P(1) is true P(1): 1 = 1(1+1)/2 = 1·2/2 = 1 ✓ TRUE STEP 2 (ASSUME): Assume P(k) is true P(k): 1 + 2 + 3 + ... + k = k(k+1)/2 ... (i) STEP 3 (PROVE): Show P(k+1) is true We n…

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STEP 1 (BASE CASE): P(1) true? P(1): 1 = 1² = 1 ✓ TRUE STEP 2 (ASSUME): Assume P(k) is true P(k): 1 + 3 + 5 + ... + (2k-1) = k² ... (i) STEP 3 (PROVE): Show P(k+1) is true We need: 1 + 3 + ... + (2k…