Answer: 2k >= k+1 for k >= 1.
- A 2k >= k+1 for k >= 1
- B k greater than 0, with no other condition needed
- C 2<sup>k</sup> less than k, an inequality that rarely holds
- D k is an even natural number specifically
Correct answer: A. 2k >= k+1 for k >= 1
Explanation: Since 2<sup>k</sup> > k by hypothesis, 2 x 2<sup>k</sup> > 2k. To conclude 2<sup>k+1</sup> > k+1, we need 2k >= k+1, which holds for all k >= 1.
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.