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This is question 7a in Section 2.1 in Pommersheim's Number Theory: A Lively Introduction With Proofs, Applications, and Stories.

Prove by mathematical induction that $\forall n\in \Bbb N,~\frac{n(n+1)(n+2)}{6}\in \Bbb N$

The work I have so far:

$P(1): \frac{1(1+1)(1+2)}{6} = \frac{6}{6} = 1$

$1\in \Bbb N$

$P(k): \frac{k(k+1)(k+2)}{6}\in \Bbb N , k\in \Bbb N$

$P(k+1): \frac{(k+1)(k+2)(k+3)}{6}\in \Bbb N , k\in \Bbb N$

$\frac{(k^2+3k+2)(k+3)}{6}\in \Bbb N$

$\frac{(k^3+6k^2+11k+6)}{6}\in \Bbb N$

After this point, I'm not totally sure where to take the proof. Please help if you can!

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    **Hint:** $$ \frac{(k+1)(k+2)(k+3)}{6} - \frac{k(k+1)(k+2)}6 = \frac{(k+3-k)(k+1)(k+2)}6 = \frac{(k+1)(k+2)}{2} $$ – 2017-02-24
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    You could also argue directly, without induction, that one of $k$, $k+1$, and $k+2$ will be divisible by $3$, and at leas one of them will be even, so you can always divide $2\cdot 3$ out from their product. – 2017-02-24
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    Another possible way: We want to use the ind hypo, so we split the term in this way: $\frac{(k + 1)(k + 2)(k + 3)}{6} = \frac{k(k + 1)(k + 2)}{6} + \frac{3(k + 1)(k + 2)}{6} = \frac{k(k + 1)(k + 2)}{6} + \frac{(k + 1)(k + 2)}{2}$. Now notice one of $k + 1$ and $k + 2$ is even. – 2017-02-24
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    Out of three consecutive numbers one is divisible by $3$ while at lest one is even, i.e., it's divisible by $2$. Therefore the whole term is divisible by $6$. – 2017-02-24
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    @TylerScott When doing induction, one scratch their head to use the inductive hypothesis. I remember it was hard when I first learned about it, but it is a very useful technique. Recently, I use induction on degree to prove a statement involving polynomial when doing exercise in analysis, which differs from the standard analytic proof. It is very satisfying when it works! – 2017-02-24

5 Answers 5

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you know that $P(k) \in \mathbb{N}$

thus $P(k+1) = P(K) + \frac{3(k+1)(k+2)}{6} = P(k) + \frac{(k+1)(k+2)}{2}$

and the latter term is in$\mathbb{N}$ of course.

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$$\text{Check }P(k+1): \frac{(k+1)(k+2)(k+3)}{6}, k\in \Bbb N$$

$$= \frac{(k+1)(k+2)k}{6} + \frac{3(k+1)(k+2)}{6}, k\in \Bbb N$$

$$\implies \underbrace{\frac{(k+1)(k+2)k}{6}}_{\in \Bbb Z\ by\ P(k)} + \underbrace{\frac{(k+1)(k+2)}{2}}_{\in \Bbb Z \text{ since one is odd, another one is even}} \in \Bbb N , k\in \Bbb N \tag{*}\label{*}$$

Edit in response to a comment by @HenningMakholm

Since the argument type "one is odd, another one is even" (in fact taking $(k+1)(k+2) \pmod 2$) can immediately kill the question (by taking $(k+1)(k+2)(k+3) \pmod 6$), to avoid such kind of argument, we establish another "induction". I write it out explicitly hoping that others understand this.

$\forall n\in \Bbb N, Q(n): \frac{n(n+1)}{2}\in \Bbb N$

We have $Q(1) = \frac{1 \cdot 2}{2} = 1$, and we assume $Q(k): \frac{k(k+1)}{2}\in \Bbb N$.

$$Q(k+1): \frac{(k+1)(k+2)}{2} = \frac{k(k+1) + 2(k+1)}{2} \\ = \underbrace{\frac{k(k+1)}{2}}_{\in \Bbb N \text{ by } Q(k)} + (k+1) \in \Bbb N$$

We use $Q(n)$ to show that the second term in \eqref{*} is in $\Bbb N$. In this way, we have a "proof completely by induction" without the kind of "overkilling arguments" spotted out in the comment below.

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    The "one is odd, another one is even" style of argument could also have been used directly to prove the original goal without any overt induction. – 2017-02-24
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    @HenningMakholm You're right, but the OP asks for a "proof by induction". – 2017-02-24
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    True -- I'm just suggesting that for that constraint to make sense one would probably want to prove that $\binom {k+2}2$ is an integer by a _separate_ inductive proof (or referring back to an earlier exercise, if available). – 2017-02-24
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    @HenningMakholm Edited my post in response to your comment in order to avoid arguments involving "taking mod", despite the tag ([tag:elementary-number-theory]) for this question. – 2017-02-24
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Inductive Step:

We should have $P (k+1)-P (k)$ to be a natural number if both these expressions belong to $\mathbb N $. Let us check that out.

We get, $$P (k+1)-P (k) = \frac {(k+1)(k+2)(k+3)}{6} - \frac {k (k+1)(k+2)}{6} = \frac {(k+1)(k+2)}{6}[k+3-(k)] $$ $$=\frac {(k+1)(k+2)}{2} $$ which is a natural number for all $k $ (why ? Note that it is the sum of the first $k+1$ natural numbers)

So, as $P (k+1) $ is true, the proposition has been proved to be true. Hope it helps.

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You should not develop the expression, and I direct reasoning with no induction is possible here.

Let me propose you another way to prove your result.

You know since $n$ and $(n+1)$ are two consecutive numbers that $n(n+1)$ is a multiple of $2$. So $n(n+1)(n+2)$ is also a multiple of $2$.

And $n$, $(n+1)$ and $(n+2)$ are three consecutive numbers, so $n(n+1)(n+2)$ is a multiple of $3$.

And because $\gcd(2,3)=1$:

$$n(n+1)(n+2)$$

is a multiple of $6$.

So

$$\frac{n(n+1)(n+2)}6$$

is an integer.

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    However, since the point of the exercise is probably to train the use of induction rather than learning the result, this won't really help the OP get there. – 2017-02-24
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Other induction:

Among $n,n+1,n+2$, one is even and one is a multiple of $3$ (so that the product is a multiple of $6$) implies that among $n+1,n+2,n+3$, one is even and one is a multiple of $3$.

Indeed, $n$ (resp. $n+1,n+2$) is even implies that $n+2$ (resp. $n+3,n+2$) is even and $n$ (resp. $n+1,n+2$) is a multiple of $3$ implies that $n+3$ (resp. $n+1,n+2$) is a multiple of $3$.

Obviously, among $0,1,2$ one is even and one is a mutiple of $3$.

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    (Same as Henning Makholm's comment.) – 2017-02-24