Prove by induction that: n(n^2 + 3n+2) is divisible by 6

Prove by induction that: $\displaystyle n(n^2 + 3n+2) $ is divisible by 6, for n>0

For n =1,

1(1^2 + 3x1 + 2) = 6

true for n =1

assume true for n = k

$\displaystyle k (k^2 + 3k+2) = 6M $

For n = k + 1,

$\displaystyle k [(k+1)^2 + 3(k+1) + 2] $should be divisible by 6

LHS = $\displaystyle k [(k+1)^2 + 3(k+1) + 2] $

= $\displaystyle (k+1) (k^2+5k+6) $

= $\displaystyle k(k^2+5k+6) + k^2 + 5k+6 $

= $\displaystyle k(k^2 + 3k +2) + 2k^2 + 4k + k^2 + 5k + 6 $

= $\displaystyle 6M + 3k^2 + 9k + 6 $

= $\displaystyle 6M + 3(k^2 + 3k +2)$

= $\displaystyle 6M + 3(6M/k)$

= $\displaystyle 6(M+3/k)$

now I'm stuck. That is obviously not divisible by 6 because M + 3/k is not an integer

HELP

Thank you in advance