2. The function $f(x)$ is defined on $\mathbf{R}$ as an odd function, and for any $x \in \mathbf{R}$, we have
$$
f(x+10)=f(x)+f(5-x) .
$$

If $f(5)=0$, then the value of $f(2005)$ is $(\quad)$.
(A) 2000 (B) 2005 (C) 2008 (D) 0
2.D.

From the problem, we have $f((5-x)+10)=f(5-x)+f(x)$.
Therefore, $f(x+10)=f(15-x)=-f(x-15)$.
Thus, $f(x)=-f(x-25)=f(x-50)$.
Hence, $f(x)$ is a periodic function with a period of 50.
Therefore, $f(2005)=f(50 \times 40+5)=f(5)=0$.