TITLE: Prove that the quantity is an integer
QUESTION [5 upvotes]: I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?

REPLY [4 votes]: Let's continue from your last step.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+2\frac{k^2}{2}+\frac{k}{6}$$
$$=\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6}+k^2$$
By induction hypothesis, first three terms yileds an integer and $k^2$ is an integer.
Thus finally we have showed that $$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}$$ is an integer.