Answer: Because k(k+1) is a product of consecutive integers, so it is always even, making 3k(k+1) divisible by 6.
- A Because k tends to be even in many natural number cases, independent of the value of k+1
- B Because k(k+1) is a product of consecutive integers, so it is always even, making 3k(k+1) divisible by 6
- C Because 3 itself is divisible by 6 without remainder, which some take as sufficient reasoning on its own
- D Because k+1 happens to be divisible by 3 in this case, regardless of which natural number k is chosen
Correct answer: B. Because k(k+1) is a product of consecutive integers, so it is always even, making 3k(k+1) divisible by 6
Explanation: Among any two consecutive integers k and k+1, one is always even, so k(k+1) is even. Multiplying by 3 gives a number divisible by both 2 and 3, hence by 6.
Mathematical induction works like a row of dominoes: proving the base case P(1) tips the first domino, and proving the inductive step (P(k) ⟹ P(k+1)) guarantees each domino knocks over the next - together these two facts guarantee ALL dominoes fall, without checking each one individually.
Concept context
A proof technique used to establish that a statement is true for every natural number, using a base case and an inductive step.