2. For the geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=1536$, and the common ratio is $q=-\frac{1}{2}$. Let $\pi_{n}$ represent the product of its first $n$ terms, then the largest $\pi_{n}(n$ $\in \mathbf{N}$ ) is
( A) $\pi_{9}$
( B) $\pi_{11}$
(C) $\pi_{12}$
(D) $\pi_{13}$
(C)
2.【Analysis and Solution】For the geometric sequence $\left\{a_{n}\right\}$, the general term formula is $a_{n}=1536 \times\left(-\frac{1}{2}\right)^{n-1}$, and the product of the first $n$ terms is $\pi_{n}=1536^{n} \times\left(-\frac{1}{2}\right)^{\frac{n(n-1)}{2}}$. It is easy to see that $\pi_{9}, \pi_{12}, \pi_{13}$ are positive, while $\pi_{10}, \pi_{11}$ are negative. Therefore, we only need to compare $\pi_{9}, \pi_{12}, \pi_{13}$.
First, consider the sign. Clearly, $\pi_{9}, \pi_{12}, \pi_{13}$ are positive. And
$$
\begin{array}{l}
a_{10}=1536 \times\left(-\frac{1}{2}\right)^{9}=-3, a_{11}=a_{10}\left(-\frac{1}{2}\right)=\frac{3}{2}, \\
a_{12}=a_{11} \cdot\left(-\frac{1}{2}\right)=-\frac{3}{4}, a_{13}=a_{12} \cdot\left(-\frac{1}{2}\right)=\frac{3}{8},
\end{array}
$$

and $a_{10} \cdot a_{11} \cdot a_{12}=(-3)\left(\frac{3}{2}\right)\left(-\frac{3}{4}\right)=\frac{27}{8}>1$.
Thus, $\pi_{12}=a_{10} a_{11} a_{12} \pi_{9}=(-3)\left(\frac{3}{2}\right)\left(-\frac{3}{4}\right) \pi_{9}=\frac{27}{8} \pi_{9}$.
Therefore, $\pi_{12}>\pi_{9}$, and since $0<a_{13}<1$, then $\pi_{13}<\pi_{12}$.
Hence, the answer is (C).