5. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{1}=0, \\
a_{n+1}=a_{n}+2+2 \sqrt{1+a_{n}}(n=1,2, \cdots) .
\end{array}
$$

Then $a_{2009}=(\quad$.
(A) 4036080
(B) 4036078
(C) 4036082
(D) 4036099
5. A.

From the given information, we have
$$
\begin{array}{l}
a_{n+1}+1=a_{n}+1+2 \sqrt{1+a_{n}}+1 \\
=\left(\sqrt{a_{n}+1}+1\right)^{2} .
\end{array}
$$

Since $a_{n+1}>0$, then
$$
\sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+1 \text { . }
$$

Therefore, the sequence $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1, i.e., $\sqrt{a_{n}+1}=n$.
Thus, $a_{n}=n^{2}-1=(n-1)(n+1)$.
Hence, $a_{2009}=2008 \times 2010=4036080$.