Final answer:
To prove the proposition P(n) for the sum of the first n positive integers, we use Mathematical Induction. First, we verify the base case for n=1, then assume P(k) is true for an arbitrary k, and prove P(k+1) follows. This completes the proof for all positive integers n.
Step-by-step explanation:
To prove the proposition P(n) using Mathematical Induction, we want to show that for any positive integer n, the sum of the first n positive integers equals ½ n(n+1). The proof is done in two steps: the base case and the inductive step.
Base Case
First, we must show that P(1) holds true. For n = 1, the sum of the first 1 positive integers is simply 1:
1 = ½ × 1 × (1+1) = ½ × 2 = 1,
thus the base case is true.
Inductive Step
For the inductive step, we assume that P(k) is true for some arbitrary positive integer k; that is,
1 + 2 + 3 + … + k = ½ k(k+1),
and we need to show that P(k+1) follows:
1 + 2 + 3 + … + k + (k+1) = ½ (k+1)((k+1)+1).
Starting with the left side of P(k+1), we add (k+1) to both sides of the assumed P(k):
1 + 2 + … + k + (k+1) = ½ k(k+1) + (k+1).
Factoring out (k+1) on the right side gives us:
1 + 2 + … + k + (k+1) = (k+1)(½ k + 1) = ½ (k+1)(k+2),
which is exactly the expression ½ (k+1)((k+1)+1), thus completing the inductive step and proving the proposition P(n) to be true for all positive integers n.