Prove that the quantity is an integer

Yes, we can. If $f(n)=\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}$, then $f(n+1)-f(n)=n^2$.

If you want to use induction you want to show that $$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}$$ is an integer. You can expand all the terms and use the known fact about the expression for $k$ to get there.

Another approach is to note that $6$ is a common denominator and say you want to prove that the numerator $2k^3-3k^2+k$ is divisible by $6$. But $2k^3-3k^2+k=(2k-1)(k-1)k$ and one of $k$ or $k-1$ is even and one of the terms is a multiple of $3$.

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.