Find the formula for the sequence $(a_n)$ that satisfies the recurrence relation $a_n=(n+7)a_{n-1}+n^2$ with $a_0=1$ 
Find the formula for the sequence $(a_n)$ that satisfies the
  recurrence relation $a_n=(n+7)a_{n-1}+n^2$ with the initial condition
  $a_0=1$.

This is a non-linear nonhomogeneous recurrence relation. What I can think that may help solve this problem is using the idea of generating function.
Let $G(x)=\sum_{n=0}^{\infty}a_nx^n$. Then
$$\begin{aligned}
G(x)&=a_0+\sum_{n=1}^{\infty}a_nx^n\\
&=1+\sum_{n=1}^{\infty}\left((n+7)a_{n-1}+n^2\right)x^n\\
&=1+\sum_{n=1}^{\infty}\left((n+7)a_{n-1}x^n\right)+\sum_{n=1}^{\infty}(n^2x^n)
\end{aligned}$$
Then I was stuck here, because I couldn't change the two summations into forms of $G(x)$.
Anyone has brillant ideas?

Since the solution of
$$ b_{n} = (n+7) b_{n-1},\quad b_0=1 $$
is clearly given by $b_n=\frac{(n+7)!}{7!}$, it looks like a good idea to enforce the substitution $a_n=\frac{(n+7)!}{7!}A_n$, leading to $A_0=1$ and
$$ A_n - A_{n-1} = \frac{7!n^2}{(7+n)!}.\tag{A} $$
By summing both sides of $(A)$ over $n=1,2,\ldots,N$ we get
$$ A_N = 1+\sum_{n=1}^{N}\frac{7!n^2}{(7+n)!}\tag{B} $$
and
$$ a_N = \frac{(N+7)!}{7!}+\sum_{n=1}^{N}\frac{(N+7)!n^2}{(n+7)!}.\tag{C} $$
In particular, for large values of $N$ we have
$$ a_N \approx (N+7)!\cdot\left(37e-\frac{21121}{210}\right). \tag{D}$$
