<s>[INST] Prove that the quantity is an integer 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?
 [/INST] If you put everything over a common denominator you get $$f(n)=\frac {2n^3-3n^2+n}6=\frac {n(n-1)(2n-1)}6$$
Now it is easier, I think, to see what is going on for a straightforward induction, or you can write $2n-1=2n-4+3$ and split up into different fractions viz $$f(n)=\frac {n(n-1)(2n-4+3)}6=\frac {n(n-1)(n-2)}3+\frac {n(n-1)}2$$ and this is easily the sum of two integers since the product of $r$ successive integers is divisible by $r$ (easily proved by induction), or you can do the induction on $n$ based on this form of $f(n)$, which will should work easily.
</s>