1. **State the problem:** We want to prove by mathematical induction that the expression $$P_n = n(n^2 + 5)$$ is divisible by 6 for all integers $$n \geq 1$$. 2. **Base case:** Check for $$n=1$$: $$P_1 = 1(1^2 + 5) = 1 \times 6 = 6$$, which is divisible by 6. 3. **Inductive hypothesis:** Assume that for some $$k \geq 1$$, $$P_k = k(k^2 + 5)$$ is divisible by 6. That is, $$6 \mid k(k^2 + 5)$$. 4. **Inductive step:** We need to prove that $$P_{k+1} = (k+1)((k+1)^2 + 5)$$ is divisible by 6. Expand $$P_{k+1}$$: $$P_{k+1} = (k+1)(k^2 + 2k + 1 + 5) = (k+1)(k^2 + 2k + 6)$$ Distribute: $$P_{k+1} = (k+1)(k^2 + 2k + 6) = (k+1)(k^2 + 2k) + 6(k+1)$$ Rewrite: $$P_{k+1} = (k+1)k^2 + 2(k+1)k + 6(k+1) = k^2(k+1) + 2k(k+1) + 6(k+1)$$ Factor $$k(k+1)$$: $$P_{k+1} = k(k+1)(k + 2) + 6(k+1)$$ By the inductive hypothesis, $$k(k^2 + 5)$$ is divisible by 6, and note that $$k(k+1)(k+2)$$ is the product of three consecutive integers, which is always divisible by 6. Since $$k(k+1)(k+2)$$ is divisible by 6 and $$6(k+1)$$ is clearly divisible by 6, their sum $$P_{k+1}$$ is divisible by 6. 5. **Conclusion:** By mathematical induction, $$P_n = n(n^2 + 5)$$ is divisible by 6 for all $$n \geq 1$$.