I wanted to prove that $n(n+1)(2n+1)$ is always divisible by three, for this I used the following approach:
$n = 0, 1\pmod2$
for $n = 0$; $n(n+1)(2n+1) \equiv 0\pmod3$
for $n = 1$; $n(n+1)(2n+1) \equiv 0\pmod3$
But now I am unsure if this the correct way of doing because of the following eg:
Let $k = \text{an odd number}\equiv 1\mod3$
for $k = 1$; $2k + 1 \equiv 0\pmod3$, therefore for every odd number $2k + 1$ is divisible by $3$, but this is not true.
So my question is the approach I am using to prove $n(n+1)(2n+1)$ divisible by $3$ is wrong or only the second example is wrong?, and why?