We are tasked with proving by mathematical induction that $n^3 + 5n$ is divisible by 6 for all $n \in \mathbb{N}$ (natural numbers).

Steps of Mathematical Induction

We will follow the standard steps of mathematical induction:

1. Base case: Prove that the statement holds for the initial value $n = 1$.
2. Inductive hypothesis: Assume that the statement is true for some arbitrary $n = k$. That is, assume that $k^3 + 5k$ is divisible by 6.
3. Inductive step: Prove that if the statement is true for $n = k$, it is also true for $n = k + 1$.

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1. Base Case: $n = 1$

Substitute $n = 1$ into the expression:

$n^3 + 5n = 1^3 + 5(1) = 1 + 5 = 6$

Since $6$ is divisible by 6, the base case holds.

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2. Inductive Hypothesis

Assume that for some arbitrary $n = k$, the statement is true. That is, we assume:

$k^3 + 5k \text{ is divisible by 6.}$

In other words, there exists some integer $m$ such that:

$k^3 + 5k = 6m$

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3. Inductive Step

Now, we must prove that the statement is true for $n = k + 1$. That is, we need to show that:

$(k + 1)^3 + 5(k + 1) \text{ is divisible by 6.}$

Let’s expand the expression $(k + 1)^3 + 5(k + 1)$:

$(k + 1)^3 = k^3 + 3k^2 + 3k + 1$ $5(k + 1) = 5k + 5$

Now add these two results:

$(k + 1)^3 + 5(k + 1) = k^3 + 3k^2 + 3k + 1 + 5k + 5 = k^3 + 3k^2 + 8k + 6$

We can rearrange the terms:

$k^3 + 5k + 3k^2 + 6$

Now, from our inductive hypothesis, we know that $k^3 + 5k$ is divisible by 6. That is:

$k^3 + 5k = 6m \quad \text{(for some integer } m\text{)}.$

Thus, the expression becomes:

$6m + 3k^2 + 6 = 6(m + k^2 + 1)$

Since the entire expression is divisible by 6, we conclude that $(k + 1)^3 + 5(k + 1)$ is divisible by 6.

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Conclusion

Since the base case holds and the inductive step is true, by the principle of mathematical induction, we have proven that $n^3 + 5n$ is divisible by 6 for all $n \in \mathbb{N}$.

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Would you like more details on any part of the proof, or do you have any questions?

Here are 5 related questions for further exploration:

1. How would the proof change if we needed to prove divisibility by a different number, say 4 or 8?
2. What would happen if we tried proving divisibility by 6 for the expression $n^2 + 3n$ instead?
3. How can mathematical induction be applied to prove divisibility properties for sums or products of numbers?
4. What are other types of induction, such as strong induction, and when are they needed?
5. Can the principle of mathematical induction be used for proofs involving inequalities?

Tip: When working with induction, always check the base case carefully and ensure the algebra is correct in the inductive step. Small mistakes in expanding terms can make the proof fail.