<s>[INST] Prove by induction that $\frac{n^3}{3}+\frac{2n}{3}$ is an integer. The question that I am working on is:  

Prove that $\dfrac{n^3}{3}+\dfrac{2n}{3} \in \mathbb Z \ \forall \ n \in \mathbb N$

The method that I think would be will work for this question is that I say that $3|(n^3+2n)$ and prove that.Would this be a good way to do this question?
So far I have done the following:
1) Base Case
n = 1

$3|n^3+2n = 3|3  \checkmark$


2) Assume, $n^3+2n$ is divisible by $3$ for $n =k, k \in \mathbb N$


$k^3+2k = 3m , m \in \mathbb N$


Let $n = k + 1$ ; Then:


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


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


$3m + 3k^2+3k+3$   
Now, I'm stuck I don't know what else to do further.
 [/INST] Look at this:
assuming
$\dfrac{k^3 + 2k}{3} \in \Bbb Z, \tag{1}$
we have
$\dfrac{(k + 1)^3 + 2(k + 1)}{3} = \dfrac{k^3 + 3k^2 + 3k + 1 + 2k + 2}{3} = \dfrac{(k^3 + 2k) + 3(k^2 + k +1)}{3}$
$= \dfrac{k^3 + 2k}{3} + k^2 + k + 1 \in \Bbb Z, \tag{2}$
by (1).
QED!!!
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