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.

  • $\begingroup$ So you have tried it for $n=1$ ? What about the induction step? What did you try? $\endgroup$ – Matti P. Mar 26 '19 at 12:17
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    $\begingroup$ 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". $\endgroup$ – Andrés E. Caicedo Mar 26 '19 at 12:18
  • $\begingroup$ The statement is that for each integer $n\geq 0$, the above sum is an integer. $\endgroup$ – Wuestenfux Mar 26 '19 at 12:18
  • $\begingroup$ 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. $\endgroup$ – Shames Mar 26 '19 at 12:19
  • $\begingroup$ 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. $\endgroup$ – 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.

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