Prove that 𝒏(𝒏𝟐 + 𝟓) is divisible by 𝟑 for any integer 𝒏 ≥ 𝟎

Sure, here is the proof:

Step 1: We need to prove that n(n^2 + 5) is divisible by 3 for any integer n ≥ 0.

Step 2: We can rewrite the expression as n^3 + 5n.

Step 3: Now, we will use the principle of mathematical induction to prove this.

Step 4: Base Case: Let's start with the smallest integer n = 0. Substituting n = 0 in the expression, we get 0^3 + 5*0 = 0 which is divisible by 3.

Step 5: Inductive Step: Assume that the statement is true for some integer k ≥ 0, i.e., k(k^2 + 5) is divisible by 3.

Step 6: We need to prove that the statement is true for the next integer k+1, i.e., (k+1)((k+1)^2 + 5) is divisible by 3.

Step 7: Expanding the expression, we get (k+1)(k^2 + 2k + 1 + 5) = k^3 + 3k^2 + 6k + 6.

Step 8: We can rewrite this as (k^3 + 5k) + 3k^2 + 6k + 6.

Step 9: By the inductive hypothesis, we know that (k^3 + 5k) is divisible by 3. Also, 3k^2 + 6k + 6 is clearly divisible by 3.

Step 10: Therefore, the sum of these two terms, i.e., (k^3 + 5k) + 3k^2 + 6k + 6, is also divisible by 3.

Step 11: Hence, by the principle of mathematical induction, n(n^2 + 5) is divisible by 3 for any integer n ≥ 0.

This completes the proof.