Final answer:
To prove that 2n < n! for n ≥ 4 using induction, we can establish a base case and then assume the inequality holds for a positive integer k and show that it also holds for k + 1.
Step-by-step explanation:
To prove that 2n < n! for n ≥ 4 using induction, we first need to establish a base case. Let's verify that the inequality holds for n = 4.
When n = 4, 2n = 2(4) = 8 and n! = 4! = 4 × 3 × 2 × 1 = 24. Since 8 is indeed less than 24, the inequality holds for the base case.
Now, let's assume that the inequality holds for some arbitrary positive integer k, where k ≥ 4. That is, we assume 2k < k!.
We need to show that the inequality also holds for k + 1. Let's start with 2(k + 1) = 2k + 2. Using the induction hypothesis, 2k < k!.
Since k! is a product of k and some positive integers less than k, we can rewrite it as k!(k + 1). Therefore, 2k! < k!(k + 1). Now, we have the following inequality: 2k + 2 < k!(k + 1).
To complete the proof, we need to show that k!(k + 1) < (k + 1)!. We can do this by dividing both sides of the inequality by (k + 1) to get k! < k!. Since k is a positive integer greater than or equal to 4, the inequality holds.