(2. For the geometric sequence $\left\{a_{n}\right\}$, the first term $a_{1}=1536$, and the common ratio $q=-\frac{1}{2}$. Let $\pi_{n}$ denote the product of its first $n$ terms, then the largest $\pi_{n}(n \in N)$ is
(A) $\pi_{9}$;
(B) $\pi_{11}$;
(C) $\pi_{12}$;
(D) $\pi_{13}$.
Answer (i)
2. (C)

The general term formula of the geometric sequence $\left\{a_{n}\right\}$ 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, and $\pi_{10}, \pi_{11}$ are negative, so we only need to compare $\pi_{9}, \pi_{12}, \pi_{13}$.
Since $a_{10}=1536 \times\left(-\frac{1}{2}\right)^{9}=-3, \quad a_{11}=a_{10} \cdot\left(-\frac{1}{2}\right)=\frac{3}{2}$,
$$
\begin{array}{l}
a_{12}=a_{11} \cdot\left(-\frac{1}{2}\right)=-\frac{3}{4} \\
a_{13}=a_{12} \times\left(-\frac{1}{2}\right)=\frac{3}{8}
\end{array}
$$

and $a_{10} \cdot a_{11} \cdot a_{12}=(-3)\left(\frac{3}{2}\right)\left(-\frac{3}{4}\right)=\frac{27}{8}>1$.
$$
\therefore \pi_{12}=a_{10} \cdot a_{11} \cdot a_{12} \cdot \pi_{9}>\pi_{9} \text {. Also, } 0<a_{13}<1 \text { and } \pi_{13}=a_{13} \cdot \pi_{12}, \therefore \pi_{13}<
$$
$\pi_{12}$, so the answer is (C).
(Indeed, for $n \geqslant 13$, $\left|a_{n}\right| \leqslant a_{13}=\frac{3}{8} . \pi_{12}$ is the maximum value of $\pi_{n}$.)