proof of divisibility of n(n+1)(2n+1) by 6

Question

How can I prove that $n(n+1)(2n+1)$ (where $n$ is a positive integer) is divisible by 6? As the product is even it is divisible by 2. But I do not know how to prove that it is divisible by 3

Answer 1

$n(n+1)(2n+1)$ is a multiple of $2$ because either $n$ or $n+1$ is even.

$n(n+1)(2n+1)=\dfrac{2n(2n+2)(2n+1)}{4}$ is a multiple of $3$ because one of $2n,2n+1,2n+2$ is a multiple of $3$ (and $4$ is not a multiple of $3$).

So $n(n+1)(2n+1)$ is a multiple of $2\times 3=6$.

Answer 2

Another (crazy) way is to see that $$1+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$

Answer 3

$$n(n+1)(2n+1)=n(n+1)(2n+4−3)$$ $$=2\cdot\underbrace{n(n+1)(n+2)}_{\text{ The product of three consecutive integers }}−3\cdot\underbrace{n(n+1)}_{\text{ The product of two consecutive integers }}$$

Now The product of n consecutive integers is divisible by n factorial