Assistance with mathematical induction to prove an expression is an integer

Question

This is my first step into this world; I'm trying my best to prove that for every $n$ this expression would be an integer: $$\frac{n}3 +\frac{n^2}2 + \frac{n^3}6.$$ I had an easier time with induction proofs when I had a series of indexes and their sum, but now I'm having some trouble proving this one. Thanks for the assistance.

Answer 1

To prove this by induction, note that it's true for $n=1$ (base case).

For the inductive step, suppose that $\frac{n}{3}+\frac {n^2}{2}+\frac{n^3}{6}$ is an integer. Then $$\frac{n+1}{3}+\frac{(n+1)^2}{2}+\frac{(n+1)^3}{6}\\=\frac{n}{3}+\frac{1}{3}+\frac{n^2}{2}+\frac{2n}2+\frac{1}{2}+\frac{n^3}{6}+\frac{3n^2}{6}+\frac{3n}{6}+\frac{1}{6}\\$$ $$=\frac{n}{3}+\frac{1}{3}+\frac{n^2}{2}+n+\frac{1}{2}+\frac{n^3}{6}+\frac{n^2}{2}+\frac{n}{2}+\frac{1}{6}\\ =\frac{n}{3}+\frac{n^2}{2}+\frac{n^3}{6}+\frac{n^2}2+\frac{3n}2+1.$$ By the inductive hypothesis, $\frac{n}{3}+\frac{n^2}{2}+\frac{n^3}{6}$ is an integer. Furthermore, $\frac{n^2}2+\frac{3n}2+1$ = $\frac{(n+1)(n+2) }2$ is an integer. This concludes the induction.

Therefore, for all $n\geq 1$, the expression is indeed an integer.