# Assistance with mathematical induction to prove an expression is an integer

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.

• So you have tried it for $n=1$ ? What about the induction step? What did you try? – Matti P. Mar 26 '19 at 12:17
• You do not prove integers, that makes no sense. Also, you do not prove that equations are integers. The word you need to use is "expression" rather than "equation". – Andrés E. Caicedo Mar 26 '19 at 12:18
• The statement is that for each integer $n\geq 0$, the above sum is an integer. – Wuestenfux Mar 26 '19 at 12:18
• I've tried n=1 following by n+1 but i cant get to a clear proof. I've also tried n=0 and going by assuming that there are "n" which are not fulfilling this equation and working with (n-1) as the minimum index. i feel like im just missing the point. – Shames Mar 26 '19 at 12:19
• What I would do, is to take $f(n) =\frac{n}{3} + \frac{n^2}{2} + \frac{n^3}{6}$ and then calculate $f(n+1)-f(n)$; and then factor the result. – Matti P. Mar 26 '19 at 12:22

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.