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?

$$F=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{n(n-1)(2n-1)}{6}$$
You can see that for any n(odd or even) the numerator is always a multiple of 6, so the sum of fractions is an integer; in fact we may have:


*

*$n=6k  ⇒ F=6m/6=m $

*$n=6k+1  ⇒ F=(6k+1)(6k+1-1)(12k+2-1)=6m/6=m$

*$n=6k+2  ⇒ F=(6k+2)(6k+2-1)(12k+4-1)=6t/6=t$

*$n=6k+3  ⇒ F=(6k+3)(6k+3-1)(12k+6-1)=6s/6=s$

*$n=6k+4  ⇒ F=(6k+4)(6k+4-1)(12k+8-1)=6v/6=v$

*$n=6k+5  ⇒ F=(6k+5)(6k+5-1)(12k+10-1)=6u/6=u$
Where m,t,s,u and v are integers.
