2. Given $\left\{a_{n}\right\}$ is a geometric sequence, $a_{2}=2, a_{5}=\frac{1}{4}$. Then the range of $a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}\left(n \in \mathbf{N}_{+}\right)$ is ( ).
(A) $[12,16)$
(B) $[8,16)$
(C) $\left[8, \frac{32}{3}\right)$
(D) $\left[\frac{16}{3}, \frac{32}{3}\right)$
2. C.

Let the common ratio of $\left\{a_{n}\right\}$ be $q$. Then $q^{3}=\frac{a_{5}}{a_{2}}=\frac{1}{8}$. Therefore, $q=\frac{1}{2}$. Hence, the sequence $\left\{a_{n} a_{n+1}\right\}$ is a geometric sequence with the first term $a_{1} a_{2}=8$ and the common ratio $q^{2}=\frac{1}{4}$. Thus,
$$
a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}=\frac{32}{3}\left(1-4^{-n}\right) .
$$

Clearly, $8=a_{1} a_{2} \leqslant a_{1} a_{2}+a_{2} a_{3}+\cdots+a_{n} a_{n+1}<\frac{32}{3}$.