lean_workbook_plus_66907

Factor into $n(n+1)(2n+1)$ . Either $n$ or $n+1$ must be even, so for it to be divisible by 6, one of the terms must be divisible by 3. The only way $n$ or $n+1$ isn't divisible by 3 is if $n\equiv 1 \pmod 3$ and $n+1 \equiv 2 \pmod 3$ . However, in this case, $2n+1 \equiv 3 \equiv 0 \pmod 3$ , so one of the three terms is always divisible by 3, and thus the expression is always divisible by 6.