$$2n^3+3n^2+7n \pmod{6} = 2n^3+3n^2+n = n(2n+1)(n+1)$$ We proceed with casework. $$n \equiv 0\pmod{6}$$where $n$ is divisible by $6$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 1\pmod{6}$$$2n+1 \equiv 2\pmod{6}$ and $n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 2\pmod{6}$$$n \equiv 2\pmod{6}$ and $n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 3\pmod{6}$$$n \equiv 3\pmod{6}$ and $n+1 \equiv 4\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 4\pmod{6}$$$n \equiv 4\pmod{6}$ and $2n+1 \equiv 3\pmod{6}$ thus $2n^3+3n^2+n$ will also be divisible by $6$ $$n \equiv 5\pmod{6}$$where $n+1$ is divisible by $6$ thus $2n^3+3n^2+n$ is also divisible by $6$ We have exhausted all cases and we have proven that it works for all of them thus we are done.

you can shorten the solution by noting that the expression is always divisible by $2$ and it is not difficult to see that the expression is always divisible by $3$

Apply induction

Let $S = 2n^3+3n^2+7n$. $S \equiv n^2+n = n(n+1) \equiv 0 \pmod 2$ $S \equiv -n^3+n = -n(n-1)(n+1) \equiv 0 \pmod 3$ Done.

Note that $2n^3+3n^2+7n\equiv 2n^3+3n^2+n\pmod 6$. We will prove that $6\mid 2n^3+3n^2+n$ for each $n$, which solves the problem. Claim: $2n^3+3n^2+n\equiv 0\pmod 2$. Note that $2n^3+3n^2+n\equiv 3n^2+n=n(3n+1)\pmod 2$. If $n$ is odd, then $3n+1$ is even, so $n(3n+1)$ is even, and if $n$ is even, we also have $n(3n+1)$ is even. $\blacksquare$ Claim: $2n^3+3n^2+n\equiv 0\pmod 3$. Note that $2n^3+3n^2+n\equiv 2n^3+n=n(2n^2+1)\pmod 3$. If $3\mid n$, then clearly $3\mid n(2n^2+1)$. If $3\nmid n$, then $n^2\equiv 1\pmod 3$, so $3\mid 2n^2+1\mid n(2n^2+1)$. $\blacksquare$ So $2n^3+3n^2+n\equiv 0\pmod 6$, as desired.

Solution: $n(2{{n}^{2}}+3n+7)=n\left( 3(n(n+1)+2)+1-{{n}^{2}} \right)$ it is enought that $6\left| (n-1)n(n+1) \right.$ and this is true, because product of 2 consecutive integer number it is divisible by 2, and from 3 too.

Notice $$2n^3+3n^2+7n\equiv 2n^3+3n^2+n\equiv n(n+1)(2n+1)\pmod{6}.$$Since $$\frac{n(n+1)(2n+1)}{6}=1^2+2^2+\dots+n^2\in\mathbb{N}$$by induction, $6\mid n(n+1)(2n+1).$ $\square$

$n \equiv 0 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 1 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 2 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 3 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 4 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ $n \equiv 5 \pmod{6} \Rightarrow 2n^3+3n^2+7n \equiv 0 \pmod{6}$ Therefore, $6|2n^3+3n^2+7n$ $\blacksquare$

Use CRT and win.