(5) Given the sequence $\left\{a_{n}\right\}$ satisfies $a_{1}=0, a_{n+1}=a_{n}+1+2 \sqrt{1+a_{n}}(n=$ $1,2, \cdots)$, then $a_{2009}=(\quad)$.
(A) 4036080
(B) 4036078
(C) 4036082
(D) 4036099
(5) A Hint: From the given, we have
$$
a_{n+1}+1=a_{n}+1+2 \sqrt{1+a_{n}}+1=\left(\sqrt{a_{n}+1}+1\right)^{2} .
$$

Since $a_{n+1}>0$, we have
$$
\sqrt{a_{n+1}+1}=\sqrt{a_{n}+1}+1,
$$

thus the sequence $\left\{\sqrt{a_{n}+1}\right\}$ is an arithmetic sequence with the first term 1 and common difference 1, $\sqrt{a_{n}+1}=n$, i.e.,
$$
a_{n}=n^{2}-1=(n-1)(n+1) .
$$

Therefore,
$$
a_{2009}=2008 \times 2010=4036080 .
$$

Hence, the answer is A.