To prove that $n^3 + 5n$ is divisible by 6 for every $n \in \mathbb{N}$, we can use mathematical induction.

Step 1: Base Case

For $n = 1$, we calculate $n^3 + 5n$: $n^3 + 5n = 1^3 + 5 \times 1 = 1 + 5 = 6$ Since 6 is divisible by 6, the base case holds true.

Step 2: Inductive Hypothesis

Assume that for some arbitrary $n = k$, the statement is true. That is, assume: $k^3 + 5k \text{ is divisible by 6}.$ This means: $k^3 + 5k = 6m \quad \text{for some integer } m.$ We need to prove that the statement holds for $n = k+1$, i.e., that $(k+1)^3 + 5(k+1)$ is divisible by 6.

Step 3: Inductive Step

We now compute $(k+1)^3 + 5(k+1)$: $(k+1)^3 + 5(k+1) = (k^3 + 3k^2 + 3k + 1) + 5k + 5$ Simplifying: $(k+1)^3 + 5(k+1) = k^3 + 3k^2 + 8k + 6$ Now, we separate terms: $(k+1)^3 + 5(k+1) = (k^3 + 5k) + 3k^2 + 6$ By the inductive hypothesis, $k^3 + 5k$ is divisible by 6, so we can write: $k^3 + 5k = 6m \quad \text{for some integer } m.$ Thus, the expression becomes: $(k+1)^3 + 5(k+1) = 6m + 3k^2 + 6$ We factor out 3 from the remaining terms: $(k+1)^3 + 5(k+1) = 6m + 3(k^2 + 2)$ Clearly, $6m + 3(k^2 + 2)$ is divisible by 6, since both terms are multiples of 6.

Conclusion:

By mathematical induction, we have shown that $n^3 + 5n$ is divisible by 6 for all $n \in \mathbb{N}$.

Would you like further details or have any questions?

Here are 5 related questions to expand the topic:

1. Can we generalize this proof to show that other polynomials are divisible by a specific number?
2. How can divisibility rules help in simplifying inductive proofs?
3. What are some common divisibility properties used in algebraic proofs?
4. Can induction be used to prove inequalities or non-divisibility properties?
5. How would this proof change if we were proving divisibility by a different number, like 3 or 12?

Tip: Always verify the base case carefully—without it, the entire inductive proof may fail.