Answer: (k+1)! = (k+1) x k!, and since k+1 > 2 for k >= 4, (k+1)! > 2 x k! > 2 x 2 k = 2 k+1.
- A (k+1)! = (k+1) x k!, and since k+1 > 2 for k >= 4, (k+1)! > 2 x k! > 2 x 2<sup>k</sup> = 2<sup>k+1</sup>
- B (k+1)! = k! + (k+1), treating factorial growth as additive rather than multiplicative
- C (k+1)! = 2 x k!, fixing the multiplier at 2 regardless of the actual value of k+1
- D (k+1)! is unrelated to k!, since each factorial is computed independently
Correct answer: A. (k+1)! = (k+1) x k!, and since k+1 > 2 for k >= 4, (k+1)! > 2 x k! > 2 x 2<sup>k</sup> = 2<sup>k+1</sup>
Explanation: By definition (k+1)! = (k+1) x k!. Since k >= 4, k+1 >= 5 > 2, so (k+1)! > 2 x k!. Using the hypothesis k! > 2<sup>k</sup>, we get (k+1)! > 2 x 2<sup>k</sup> = 2<sup>k+1</sup>.
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.