Prove that $6|2n^3+3n^2+n$ My attempt at it: $\displaystyle 2n^3+3n^2+n= n(n+1)(2n+1) = 6\sum_nn^2$
This however reduces to proving the summation result by induction, which I am trying to avoid as it provides little insight.

to check for divisibility by 6 a number must be divisible by both 2 and 3 so we will prove that
$2n^3 + 3n^2 + n
= n (2n^2 + 3n +1)
= n (n+1) (2n+1)$
If $n$ is even then 2 divides $n$ and $n+1$ will be odd so $n+1$ can be $3k+2$ or $3k$ where $k$ is some integer.
If $3k+1 = n+1$ as it would make $n$ itself a multiple of 6 as our assumption that $n$ is even
So if $n+1$ is $3k$ it can be divided by 3 so no problem.
But if $n+1 = 3k+2$ then $2n+1$ will be $2(3k+1) +1 = 6k+3$ which is divisible by 3
and hence by 6.
Case 2: If $n$ is odd then $n+1$ is even and thus divisible by 2.


*

*if $n$ is $3k$ then it is divisible by 3 so no issues

*if $n$ is $3k+2$ as it makes $n+1=3k+3$ which is itself a multiple of 6 as our earlier assumption is that $n+1$ is even

*if $n$ is $3k+1$ then $2n+1$ becomes $2(3k+1)+1=6k+3$ which is divisible by 3
and hence proved 

