Prove that for $n \geqslant 1,$ the integer $n(n+1)(2 n+1)$ is divisible by 6

Question

Prove that for $n \geqslant 1,$ the integer $n(n+1)(2 n+1)$ is divisible by $6.$

My attempt. We enough to show that $2|n(n+1)(2 n+1)$ and $3|n(n+1)(2 n+1)$. Note that any integer $n$ is of the form: $2k$ or $2k+1.$

So $n(n+1)(2n+1)=2k(2k+2)(4k+1)$. Hence $2|2k(2k+2)$, that is $2|n(n+1)$.

How can I continue this proof? Can you help?

any integer is either $3m,3m+1,3m+2$. Then proceed as when You proved it is divisible by $2$. — acat3, Apr 11 '20 at 00:35
@RezhaAdrianTanuharja Thanks for helping, I needed this too much to complete proof :)) — , Apr 11 '20 at 00:42

score 7 · Answer 1 · answered Apr 11 '20 at 00:02

7

A cool proof is given by $$n(n+1)(2n+1)=6\cdot\sum_{k=1}^n k^2.$$

answered Apr 11 '20 at 00:02

Maximilian Janisch

13,536

how can I use this? – Apr 11 '20 at 00:03
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Well $$6\cdot\sum_{k=1}^n k^2$$ is surely divisible by $6$ (and the formula can be proven straight-forwardly using induction) – Maximilian Janisch Apr 11 '20 at 00:04
@JamesEnsor $\sum k^2$ , which is the sum of 1st n squares of natural numbers, so its definitely an integer, also $n(n+1)(2n+1)$ is clearly an integer – Saket Gurjar Apr 11 '20 at 00:08
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@MaximilianJanisch the proof is indeed "cool" – Saket Gurjar Apr 11 '20 at 00:08
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Thanks for comments and helping, the proof is very cool! :) – Apr 11 '20 at 00:08
To be honest I don't think my proof should be taken too seriously but I just couldn't not post it – Maximilian Janisch Apr 11 '20 at 00:21

score 1 · Accepted Answer · answered Apr 11 '20 at 00:19

I'd say $95\%+$ of these sorts of statements can be done by induction. So when you don't have a clue, go with that as a starter. Maximilian's proof is very elegant, but if you can't spot the identity it's difficult to replicate

Firstly, verify the base case: $n=1$. Here we have $1\cdot 2\cdot 3=6$. Clearly this is divisible by $6$.

We then assume it true for $n=k$. That is: $$k(k+1)(2k+1)=6\lambda\tag 1$$ and use this to show it true for $k+1$, i.e. that $$(k+1)(k+2)(2k+3)=6\mu\tag 2$$

Expanding $(1)$, we see $2k^3+3k^2+k=6\lambda$

and expanding $(2)$ gives $2k^3+9k^2+13k+6=6\mu$

Substitute in the $6\lambda$ into the expansion of $(2)$ and see what you get.

To be honest I don't think my proof should be taken too seriously but I just couldn't not post it — Maximilian Janisch, Apr 11 '20 at 00:21

score 1 · Answer 3 · answered Apr 11 '20 at 00:24

You can use the expansion of a polynomial using binomial coefficients, this is very useful for questions like prove $P(n)$ is divisible by $k$

$$n(n+1)(2n+1)=12\binom{n}{3}+18\binom{n}{2}+6\binom{n}{1}$$

Since binomial coefficients are integers, this is divisible by $6$.

Prove that for $n \geqslant 1,$ the integer $n(n+1)(2 n+1)$ is divisible by 6

3 Answers3