Proof ∀n∈ℕ, 3|(n³+2n)

Prove by induction that, ∀n∈ℕ, 3|(n³+2n).

We are asked to prove by mathematical induction that:

$$ \forall n \in \mathbb{N},\ 3 \mid (n^3 + 2n) $$

That is, for every natural number $n$, the expression $n^3 + 2n$ is divisible by 3.

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Step 1: Base Case

Let $n = 1$:

$$ n^3 + 2n = 1^3 + 2(1) = 1 + 2 = 3 $$

Since $3 \mid 3$, the base case holds.

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Step 2: Inductive Hypothesis

Assume that the statement holds for some $n = k$, i.e., assume:

$$ 3 \mid (k^3 + 2k) $$

This means there exists an integer $m$ such that:

$$ k^3 + 2k = 3m $$

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Step 3: Inductive Step

We want to show that the statement holds for $n = k + 1$, i.e., show:

$$ 3 \mid ((k+1)^3 + 2(k+1)) $$

Compute the expression:

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

Group terms:

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

By the inductive hypothesis, $3 \mid (k^3 + 2k)$

Also note: $3k^2 + 3k + 3 = 3(k^2 + k + 1)$, so clearly divisible by 3.

Therefore, the entire expression is the sum of two numbers divisible by 3:

$$ (k^3 + 2k) + 3(k^2 + k + 1) \Rightarrow 3 \mid ((k+1)^3 + 2(k+1)) $$

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Conclusion

By mathematical induction, we have shown that:

$$ \forall n \in \mathbb{N},\ 3 \mid (n^3 + 2n) $$

✅ Proved.