Prove By Mathematical Induction That $\forall n\in \Bbb N,~\frac{n(n+1)(n+2)}{6}\in \Bbb N$ 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!
 A: 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. 
A: 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. 
A: $$\text{Check }P(k+1): \frac{(k+1)(k+2)(k+3)}{6}, k\in \Bbb N$$
$$= \frac{(k+1)(k+2)k}{6} + \frac{3(k+1)(k+2)}{6}, k\in \Bbb N$$
$$\implies \underbrace{\frac{(k+1)(k+2)k}{6}}_{\in \Bbb Z\ by\ P(k)} + \underbrace{\frac{(k+1)(k+2)}{2}}_{\in \Bbb Z \text{ since one is odd, another one is even}} \in \Bbb N , k\in \Bbb N \tag{*}\label{*}$$

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.
A: 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.
A: 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 $3$.
