How to show that $3k+3k^2$ is divisible by 6

Question

I need to prove that $3k+3k^2$ is divisible by $6$.

I did the following:

We can see this by considering the base case $k=1$:

$$\frac{3k+3k^2}{6}=\frac{3k}{6}+\frac{3k^2}{6}=\frac{3}{6}+\frac{3}{6}=1$$ and further for $k=2$

$$\frac{3k+3k^2}{6}=\frac{3k}{6}+\frac{3k^2}{6}=\frac{6}{6}+\frac{12}{6}=3$$

and for the case $(k+1)$: $$\frac{3(k+1)+3(k+1)^2}{6}=\frac{3(k+1)}{6}+\frac{3(k+1)^2}{6},$$ we get:

$$\frac{(k+1)}{2}+\frac{(k+1)^2}{2},$$ which is: $$\frac{1}{2}(k+2)(k+1),$$

But how can I show that the last term is always divisible by 6 for integer values of $k$? If I prove the last term is divisible by six, then the question is answered. But does that requires a second induction to be made ? Any solutions to this problem appreciated.

Thanks

It is much easier. Obviously your term $3k(k+1)$ is divisible by $3$, and $k(k+1)$ also divisible by $2$, so altogether your term divisible by $6$. — Dietrich Burde, Jun 30 '23 at 09:22
Does this answer your question? For any arbitrary integer $a$, prove that $2 \mid a \left( a+1 \right)$ and $3 \mid a \left( a+1 \right) \left( a+2 \right)$ — Bill Dubuque, Aug 02 '23 at 20:09

Arthur · Accepted Answer · 2023-06-30T11:36:53.840

To actually do the induction (which isn't the easiest way to prove this, but I still think it's fun), we need to find the old expression in the new one, and use that. We assume that $\frac{3k^2+3k}6$ is an integer, and we try as hard as we can to find $\frac{3k^2+3k}6$ as we work through the next step.

Now it is time to look at $$ \frac{3(k+1)^2+3(k+1)}6\\ =\frac{3(k^2+2k+1)+3(k+1)}6\\ =\frac{3k^2+6k+3+3k+3}6\\ =\frac{3k^2+3k+6k+6}6\\ =\frac{3k^2+3k}6+\frac{6k+6}6 $$ Here we see that we end up with an integer because the first fraction is assumed to be integer by the induction hypothesis, and the second fraction is just equal to $k+1$, clearly an integer.

Note that you are very close to a non-inductive proof in your induction step. All you have left to show is that $\frac12(k+1)(k+2)$ is an integer, which is easy, because either $k+1$ or $k+2$ must be even. (You seem to think you need to show that it is divisible by $6$, and that's not what you need. You have forgotten that you went from "$3(k+1)^2+3(k+1)$ is divisible by $6$" to "$\frac{3(k+1)^2+3(k+1)}6$ is an integer".)

score 1 · Answer 2 · answered Jun 30 '23 at 11:47

1

We have $\frac{3k+3k^2}{6}=\frac{k+k^2}{2}$.

So we only need to prove that $k+k^2$ is divisible by $2$.

Also $k+k^2=k(k+1)$, then one of $k$ and $k+1$ must be even while another is odd.

So $k+k^2$ must be even, hence proved.

answered Jun 30 '23 at 11:47

Itoz Darbien

474

score 0 · Answer 3 · answered Jul 01 '23 at 19:23

0

Since, product of $n$ consecutive integer is divisible by $n!$

Therefore,
$k(k+1)$ is divisible by $2!$ ($i.e;$ $2$)

So, $3k(k+1)$$=$$3k+3k^2$ is divisible by $6$

answered Jul 01 '23 at 19:23

Sourav Mandal

31

How to show that $3k+3k^2$ is divisible by 6

3 Answers3