Using mathematical induction show that n³ + 2n is divisible by 3, n ≥ 0.

To prove that $n^3+2n$ is divisible by 3 for all $n\ge 0$, we use mathematical induction.

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Base Case (n = 0)

Substitute $n=0$ into the expression: $$0^3+2(0)=0+0=0$$ Since 0 is divisible by 3, the statement holds for $n=0$.

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Inductive Step

Inductive Hypothesis: Assume $k^3+2k$ is divisible by 3 for some $k\ge 0$. That is, $$k^3+2k=3m\text{for some integer }m.$$

To Prove: $(k+1{)}^3+2(k+1)$ is divisible by 3.

Expand $(k+1{)}^3+2(k+1)$: $$(k+1{)}^3+2(k+1)=k^3+3k^2+3k+1+2k+2$$ $$=(k^3+2k)+3k^2+3k+3$$ By the inductive hypothesis, $k^3+2k=3m$. Substitute: $$=3m+3k^2+3k+3$$ $$=3(m+k^2+k+1)$$ This is clearly divisible by 3.

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Conclusion

By the principle of mathematical induction, $n^3+2n$ is divisible by 3 for all $n\ge 0$. Thus, the statement is proven.