MHB Proving Divisibility by 6 Using Mathematical Induction

[Monoxdifly]

Prove that n(n + 1)(n + 2) is divisible by 6 for any integer n

What I have done so far:

For n = 1
1(1 + 1)(1 + 2) = 1(2)(3) = 6(1) = 6
6 is divisible by 6 (TRUE)

For n = k
k(k + 1)(k + 2) is divisible 6 (ASSUMED AS TRUE)

For n = k + 1
(k + 1)(k + 2)(k + 3) = k(k + 1)(k + 2) + 3(k + 1)(k + 2)

What am I supposed to do from here onward?

Yes, I know that:
k(k + 1)(k + 2) is divisible by 6
(k + 1) and (k + 2) are subsequent numbers so that its product is even or divisible by 2. Because (k + 1)(k + 2) is divisible by 2 then 3(k + 1)(k + 2) is divisible by 6.
Because k(k + 1)(k + 2) is divisible by 6 and 3(k + 1)(k + 2) is divisible by 6 then k(k + 1)(k + 2) + 3(k + 1)(k + 2) is divisible by 6 (PROVEN TRUE)

However, I am not sure if that follows the standard procedure of mathematical induction. Can someone please show me the "proper" "formal" way using mathematical induction continued from the bolded part?

 

[MarkFL]

I think I would give $P_n$ as:

$$n(n+1)(n+2)=6k$$ where $$k\in\mathbb{N}$$

Now, as your induction step, I would look at adding:

$$(n+1)(n+2)(n+3)-n(n+1)(n+2)=3(n+1)(n+2)=6\sum_{j=1}^{n+1}(j)$$

to $P_n$.

 

[Monoxdifly]

How can we be sure that 3(n + 1)(n + 2) is divisible by 6 without the "theory" approach I wrote in my initial post?

 

[MarkFL]

How can we be sure that 3(n + 1)(n + 2) is divisible by 6 without the "theory" approach I wrote in my initial post?

By using the summation formula instead. Or, you could prove that $n(n+1)$ is even by induction...:)

 

[Monoxdifly]

Means we have to use double induction?

 

[MountEvariste]

How can we be sure that 3(n + 1)(n + 2) is divisible by 6 without the "theory" approach I wrote in my initial post?

Either $n$ is even or $n$ is odd. If $n$ is even, let $n=2k$ then we have $3(n+1)(n+2) = 6(k+1)(2k+1)$. If $n$ is odd then let $n=2k+1$ which gives $6(k+1)(2k+3)$.

 

[MarkFL]

Either $n$ is even or $n$ is odd. If $n$ is even, let $n=2k$ then we have $3(n+1)(n+2) = 6(k+1)(2k+1)$. If $n$ is odd then let $n=2k+1$ which gives $6(k+1)(2k+3)$.

That's likely the most straightforward way to proceed. (Yes)

 

[MountEvariste]

I also like $\displaystyle 3(n+1)(n+2) = 6 \binom{n+2}{2}$, which in a sense is your summation method in disguise.