Prove that 𝒏(𝒏𝟐 + 𝟓) is divisible by 𝟑 for any integer 𝒏 ≥ 𝟎.3. Prove that 𝟔𝒏 > 𝟓𝒏 + 𝟗 for any integer 𝒏 ≥ 𝟐.

Sure, let's break down these two problems:

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

We can prove this by induction.

Base Case: For n = 0, n(n^2 + 5) = 0(0^2 + 5) = 0, which is divisible by 3.

Inductive Step: Assume that for some k ≥ 0, k(k^2 + 5) is divisible by 3. We need to prove that (k+1)((k+1)^2 + 5) is also divisible by 3.

(k+1)((k+1)^2 + 5) = (k+1)(k^2 + 2k + 1 + 5) = (k+1)(k^2 + 5) + 2k(k+1)

We know from our inductive hypothesis that (k+1)(k^2 + 5) is divisible by 3. Also, 2k(k+1) is divisible by 3 because it's a product of two integers. Therefore, the sum of these two terms is also divisible by 3.

2. Prove that 𝟔𝒏 > 𝟓𝒏 + 𝟗 for any integer 𝒏 ≥ 𝟐.

We can prove this by induction as well.

Base Case: For n = 2, 62 = 12 > 52 + 9 = 19, which is true.

Inductive Step: Assume that for some k ≥ 2, 6k > 5k + 9. We need to prove that 6(k+1) > 5(k+1) + 9.

6(k+1) = 6k + 6 and 5(k+1) + 9 = 5k + 14.

If we subtract the second equation from the first, we get 6k + 6 - 5k - 14 = k - 8.

We know from our inductive hypothesis that 6k > 5k + 9, so k > 9. Therefore, k - 8 > 0, which means that 6(k+1) > 5(k+1) + 9.