Let $n=1990$. Find $\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}\right.$ $-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}$ ).
(1990, National High School Mathematics Competition)
Solution: Let the required expression be $A$. After transformation, we get
$$
\begin{array}{l}
A=\frac{1}{2^{n}}\left(\sqrt{3^{0}}-\sqrt{3^{2}} C_{n}^{2}+\sqrt{3^{4}} C_{n}^{4}\right. \\
-\sqrt{3^{6}} C_{n}^{6}+\cdots+\sqrt{3^{1988}} C_{n}^{1988} \\
\left.-\sqrt{3^{1990}} C_{n}^{1990} \cdot\right) \text {. } \\
\end{array}
$$

From the characteristics of the binomial expansion, we know that the expression within the parentheses is part of a binomial expansion. Therefore, we can construct the following conjugate expression for $A$:
$$
\begin{array}{l}
B=\frac{1}{2^{n}}\left(-\sqrt{3} C_{n}^{1}+\sqrt{3^{3}} C_{n}^{3}-\sqrt{3^{5}} C_{n}^{5}+\right. \\
\left.\cdots+\sqrt{3^{1987}} C_{n}^{1987}-\sqrt{3^{1989}} C_{n}^{1989}\right) \text {. Then } \\
A+B i=\frac{1}{2^{n}}(-1+\sqrt{3} i)^{n} \\
=\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} i\right)^{n} \\
=w^{1990}=w^{3 \times 663} \cdot w=w \text {. } \\
=-\frac{1}{2}+\frac{\sqrt{3}}{2} i \text {. } \\
\text { ( } w=-\frac{1}{2}+\frac{\sqrt{3}}{2} i \text { ) } \\
\text { Therefore, } A=-\frac{1}{2} \text {. } \\
\end{array}
$$

The pairing strategy is an important problem-solving strategy. Using it reasonably can optimize the problem-solving process, making complex problems simpler and easier.
(Continued on page 15).