Example 7. Let $n=1990$. Find the value of $\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots+3^{994} C_{n}^{1988}-3^{995} C_{n}^{1990}\right)$. (1990, National High School Competition)
\[
\begin{array}{l}
\quad \text { Solve } F=\frac{1}{2^{n}}\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{6}+\cdots\right. \\
\left.+3^{1994} C_{n}^{1998}-3^{995} C_{n}^{1990}\right), F^{\prime}=\frac{1}{2^{n}}\left(-\sqrt{3} C_{n}^{1}+\right. \\
\sqrt{3^{3}} C_{n}^{3}-\sqrt{3^{5}} C_{n}^{5}+\cdots+\sqrt{3^{1987}} C_{n}^{1987}-\sqrt{3^{1989}} \\
\left.\cdot C_{n}^{1989}\right) \text {. Then we have } F+i F^{\prime}=\frac{1}{2^{2}}(-1+-\sqrt{3} i)^{n} \\
=\left(-\frac{1}{2}+\cdots \frac{\sqrt[3]{2}}{2} i\right)^{n}=\omega \omega^{n}=\omega^{3 \times 863} \cdot \omega=\omega \\
=-\frac{1}{2}+\frac{\sqrt{3}}{2} i . \text { By the condition of equality of complex numbers, we get } F= \\
-\frac{1}{2} \text {, that is } \frac{1}{2^{n}} \cdot\left(1-3 C_{n}^{2}+3^{2} C_{n}^{4}-3^{3} C_{n}^{5}+\cdots+3^{994}\right. \\
\left.C_{n}^{1988}-3^{995} C_{n}^{1990}\right)=-\frac{1}{2} .
\end{array}
\]