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$ \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}$

Base Case: n=1

$ => \frac{1}{3}+\frac{1}{2}+\frac{1}{6} =1 $ therefore, it holds true

I.H:

Suppose that for some $k \in \mathbb{N}$ that

$$ \frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$

is an integer.

Then;

$ \frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{k+1}{6}$

I dont know how to proceed from here

avs
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TheGamer
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  • Not sure how you would know this off hand, but you can show that your expression is actually equal to $\sum_{i=1}^n i^2$ by induction, so much be an integer. – J.G Jul 08 '17 at 00:04
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    Hint: $;\displaystyle \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} = \frac{n+3n^2+2n^3}{6} = \frac{n(n + 1)(2 n + 1)}{6}$ – dxiv Jul 08 '17 at 00:04
  • If you use the expression dxiv gave you don't need induction. You just need to show the top is always divisible by 6 and you can use modular arithmetic for that. – Shuri2060 Jul 08 '17 at 00:05
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    Hint: Multiply out what you have, remove the $\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$ and try to simplify what you have left. – Michael Burr Jul 08 '17 at 00:05
  • You can do it by induction (what Michael Burr suggested - subtracting an integer from an integer gives an integer), or do it without induction, in which case you use modular arithmetic which is essentially Nil's answer – Shuri2060 Jul 08 '17 at 00:08
  • $\frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6} = 2 \binom{n}{3} + 3 \binom{n}{2} + n.$ – Daniel Schepler Jul 08 '17 at 00:29

5 Answers5

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Using Induction (as the OP started).

Take the expression that you have and multiply it out: \begin{align} \frac{(k+1)^3}{3}+\frac{(k+1)^2}{2}+\frac{k+1}{6} &=\frac{k^3}{3}+k^2+k+\frac{1}{3}+\frac{k^2}{2}+k+\frac{1}{2}+\frac{k}{6}+\frac{1}{6}\\ &=\left(\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}\right)+(k^2+2k)+\left(\frac{1}{3}+\frac{1}{2}+\frac{1}{6}\right). \end{align} The first term in parentheses is an integer by induction, the middle term in parentheses is a sum of products of integers, so it is an integer, and the last term is $1$ by direct computation. So the sum is an integer.

Moral of the story: Sometimes, expressions look scary or complicated, but if you dive in and give it a shot, you can reduce it to something easier.

avs
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Michael Burr
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    i.e. $,f(k!+!1) = f(k)+f(0) + g(k),$ for $g(x)\in\Bbb Z[x],,$ so $,f(k),f(0)\in\Bbb Z,\Rightarrow, f(k!+!1)\in\Bbb Z\ $ $\qquad\qquad\ $ – Bill Dubuque Jul 08 '17 at 00:56
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$$\text{Sum your terms:}\,\,\, \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\dfrac{n(n+1)(2n+1)}{6}$$ $$\text{By induction on} \,\,n\text{:}\,\,\, \sum_{k=1}^nk^2 = \dfrac{n(n+1)(2n+1)}{6}$$ Since each $k$ is an integer it follows that $k^2$ is also an integer and therefore their sum is also an integer.

Ignacio Rojas
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Or you might just note that $$ \frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6} = \frac{n(n + 1)(2 n + 1)}{6} = \frac{n(n + 1)( (n + 2) + (n-1))}{6} = \binom{n+2}{3}+\binom{n+1}{3} $$

lhf
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  • I didn't "just note that", I worked from Newton's interpolation. See https://math.stackexchange.com/questions/2350039/best-representation-of-a-polynomial-as-a-linear-combination-of-binomial-coeffici. – lhf Jul 08 '17 at 01:16
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$$\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6}=\dfrac{n}{6}(n+1)(2n+1)$$ Observe that any integer $n$ is in one of the form $6k, 6k+1, 6k+2, 6k+3, 6k+4, 6k+5.$ Check each form respectively.

Bumblebee
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  • checking for $0,1,\dots 5$ is enough – krirkrirk Jul 08 '17 at 00:14
  • It's enough to note that $n(n+1)$ is divisible by $2$, since one of two consecutive integers is even. Also $n(n+1)(2n+1)$ is divisible by $3$ since one of the three consecutive integers $2n,2n+1,2n+2$ must be a multiple of $3$. Therefore the product $n(n+1)(2n+1)$ is divisible by $2 \cdot 3=6$. – dxiv Jul 08 '17 at 00:19
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I noted that the common denominator is $6$ so that the combined numerator would be $2n^3 + 3n^2 + n$. It can only yield an integer when divided by $6$ if $2n^3 + 3n^2 + n$ is divisible by $6$.

By straightforward computation, $2n^3 + 3n^2 + n = 0, 6, 30, 84, 180, 330$ for the inputs $n = 0, 1, 2, 3, 4, 5$. So the numerator is always a multiple of $6$ and the original expression is always an integer.

You only have to go up to $n = 5$ because at $6$ the values start cycling again, as you can verify with elementary algebra. [I'm doing modular arithmetic without talking about modular arithmetic, as noted by @Shuri2060. I suppose you have to take my cycling claim on faith in the context of my answer, but it's true].

user4894
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    If you are going for that argument, perhaps make it clear you're using modular arithmetic or why you only need to compute up to $5$ – Shuri2060 Jul 08 '17 at 00:14