To apply mathematical induction to a statement, we generally follow these steps: establish a base case, assume it holds for some integer k, and then show it holds for k + 1.
Let’s denote the statement as P(n): 4n > 4n – 2 for all positive integers n.
Base Case:
For n = 1:
4(1) = 4 and 4(1) – 2 = 2. Since 4 > 2, P(1) is true.
Inductive Step:
Assume P(k) is true for some arbitrary positive integer k:
4k > 4k – 2
Now we need to show that P(k + 1) is also true:
P(k + 1): 4(k + 1) > 4(k + 1) – 2
Simplifying the left side:
4(k + 1) = 4k + 4
And the right side:
4(k + 1) – 2 = 4k + 4 – 2 = 4k + 2
So we must show:
4k + 4 > 4k + 2
Subtracting 4k from both sides leads us to:
4 > 2, which is true.
Conclusion:
Since both the base case and the inductive step hold, by mathematical induction, we can conclude that the statement 4n > 4n – 2 is true for all positive integers n.