The form of the equation allows us to solve it by using only first-order difference equations (see Remark 1).
Arrange the equation as
$$\begin{cases}[a_{n}-a_{n-1}]-[a_{n-1}-a_{n-2}]=n+1,{\quad}n=3,4,\cdots\\ a_{1}=1,\ a_{0}=0,\end{cases}$$
which is a first-order difference equation in $b_{n}:=[a_{n-1}-a_{n-2}]$ for $n=2,3,\cdots$.
So, we have
$$\begin{cases}b_{n+1}-b_{n}=n+1,{\quad}n=2,3,\cdots\\ b_{2}=1,\end{cases}$$
which implies
$$\begin{aligned}b_{n}=&1\prod_{k=2}^{n-1}1+\sum_{k=2}^{n-1}(k+1)\prod_{\ell=k+1}^{n-1}1\\ =&1+\sum_{k=2}^{n-1}(k+1)\\ =&\sum_{{\color{blue}k=1}}^{n-1}k+(n-2)\\ =&\frac{1}{2}(n^{2}+n-4),\end{aligned}$$
where we have applied Remark 2(i).
Next, we set $c_{n}:=a_{n-2}$ for $n=2,3,\cdots$, then $c_{n+1}-c_{n}=a_{n-1}-a_{n-2}=b_{n}$, i.e.,
$$\begin{cases}c_{n+1}-c_{n}=\frac{1}{2}(n^{2}+n-4),{\quad}n=2,3,\cdots\\ c_{2}=0\end{cases}$$
whose solution is
$$\begin{aligned}c_{n}=&0\prod_{k=2}^{n-1}1+\frac{1}{2}\sum_{k=2}^{n-1}(k^{2}+k-4)\prod_{\ell=k+1}^{n-1}1\\ =&\frac{1}{2}\sum_{k=2}^{n-1}(k^{2}+k-4)\\ =&\frac{1}{2}\sum_{{\color{blue}k=1}}^{n-1}k^{2}+\frac{1}{2}\sum_{{\color{blue}k=1}}^{n-1}k-2(n-2)-{\color{blue}1}\\
=&\frac{1}{2}\frac{1}{6}(n-1)n(2n-1)+\frac{1}{2}\frac{1}{2}(n-1)n-2(n-2)-1\\ =&\frac{1}{6}(n^{3}-13n+18),\end{aligned}$$
where we have applied Remark 2(ii).
As $a_{n}=c_{n+2}$, we get
$$a_{n}=\frac{1}{6}(n^{3}+6n^{2}-n),{\quad}n=2,3,\cdots.\tag*{$\blacksquare$}$$
Remark 1. The first-order difference equation
$x_{n+1}-p_{n}x_{n}=q_{n}$
has the solution
$\displaystyle x_{n}=x_{m}\prod_{k=m}^{n-1}q_{k}+\sum_{k=m}^{n-1}q_{k}\prod_{\ell=k+1}^{n-1}p_{\ell}$
with the convention that the empty product is unity.
Remark 2. The following identities hold.
(i) $\displaystyle\sum_{k=1}^{m}k=\frac{1}{2}m(m+1)$.
(ii) $\displaystyle\sum_{k=1}^{m}k^{2}=\frac{1}{6}m(m+1)(2m+1)$.
