explanation ; To prove that 3 divides n3 + 2n for any positive integer n, we can use mathematical induction.
Base case: When n = 1, we have 13 + 2(1) = 3, which is divisible by 3.
Inductive step: Assume that for some positive integer k, 3 divides k3 + 2k. We want to show that 3 divides (k+1)3 + 2(k+1).
Expanding the left side, we get:
(k+1)3 + 2(k+1) = k3 + 3k2 + 3k + 1 + 2k + 2
Simplifying, we get:
(k+1)3 + 2(k+1) = (k3 + 2k) + 3k2 + 3k + 3
By the inductive hypothesis, k3 + 2k is divisible by 3. Also, 3k2 + 3k is divisible by 3 because it has a common factor of 3. Therefore, (k+1)3 + 2(k+1) is divisible by 3.
Since the base case holds and the inductive step shows that if 3 divides k3 + 2k, then 3 divides (k+1)3 + 2(k+1), we can conclude that 3 divides n3 + 2n for any positive integer n.