Example 6 (1990 National High School Mathematics League Question) Let $n=1990$, find $\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots\right.$ $\left.+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}\right)$
Let the required expression be $A$. After transformation, we get
$$
A=\frac{1}{2^{n}}\left(\sqrt{3^{0}}-\sqrt{3^{2}} C_{n}^{2}+\sqrt{3^{4}} C_{n}^{4}-\sqrt{3^{6}} C_{n}^{6}+\cdots+\sqrt{3^{1988}} C_{n}^{1988}-\sqrt{3^{1990}} C_{n}^{1990}\right) \text {. }
$$

By the characteristics of the binomial expansion, the expression within the parentheses is part of a binomial expansion. Therefore, we can construct the following conjugate expression for $A$:
$$
B=\frac{1}{2^{n}}\left(-\sqrt{3} C_{n}^{1}+\sqrt{3^{3}} C_{n}^{3}-\sqrt{3^{5}} C_{n}^{5}+\cdots+\sqrt{3^{1987}} C_{n}^{1987}-\sqrt{3^{1989}} C_{n}^{1989}\right),
$$

Then $A+B \mathrm{i}=\frac{1}{2^{n}}(-1+\sqrt{3} \mathrm{i})^{n}$
$$
=\left(-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i}\right)^{n}=\omega^{1990}=\omega^{3 \times 663} \cdot \omega=\omega=-\frac{1}{2}+\frac{\sqrt{3}}{2} \mathrm{i} \text {. }
$$

Therefore, $A=-\frac{1}{2}$.