3. The sequence $\left\{a_{n}\right\}$ satisfies
$$
a_{n}=\frac{5 a_{n-1}-2}{a_{n-1}-5}\left(n \geqslant 2, n \in \mathbf{N}_{+}\right),
$$

and the sum of the first 2008 terms is 401. Then the value of $\sum_{i=1}^{2008} a_{i} a_{i+1}$ is ( ).
(A) -4
(B) -6
(C) 6
(D) 4
3. B.

Let $a_{1}=m$, then we have
$$
a_{2}=\frac{5 m-2}{m-5}, a_{3}=\frac{5 a_{2}-2}{a_{2}-5}=m=a_{1} .
$$

It is easy to see that the period of $a_{n}$ is 2.
Thus, $a_{2009}=a_{2007}=\cdots=a_{1}$.
Notice that $a_{n+1}=\frac{5 a_{n}-2}{a_{n}-5}$, so,
$$
\begin{array}{l}
a_{n} a_{n+1}=5\left(a_{n}+a_{n+1}\right)-2 . \\
\text { Therefore, } \sum_{i=1}^{2008} a_{i} a_{i+1}=5 \sum_{i=1}^{2008}\left(a_{i}+a_{i+1}\right)-2 \times 2008 \\
=10\left(a_{1}+a_{2}+\cdots+a_{2008}\right)-4016=-6 .
\end{array}
$$