2024/036) Show that $6∣2n^3+3n^2+n,n\in N$

 we have

$2n^3+3n^2 + n = n(2n^2+3n +1) = n(2n+1)(n+1)$

$= n((n-1) + (n+2))(n+1) = (n-1)n(n+1) + n(n+1)(n+2)$

$(n-1)n(n+1)$ is product of 3 consecutive integers and hence divisible by 6 and so is $n(n+1)(n+2)$ and hence the sum is divisible by 6