28. Find the value of
$$
\binom{2003}{1}+\binom{2003}{4}+\binom{2003}{7}+\cdots+\binom{2003}{2002} .
$$
Solution: $\quad\left(2^{2003}-2\right) / 3$
Let $\omega=-1 / 2+i \sqrt{3} / 2$ be a complex cube root of unity. Then, by the binomial theorem, we have
$$
\begin{aligned}
\omega^{2}(\omega+1)^{2003} & =\binom{2003}{0} \omega^{2}+\binom{2003}{1} \omega^{3}+\binom{2003}{2} \omega^{4}+\cdots+\binom{2003}{2003} \omega^{2005} \\
2^{2003} & =\binom{2003}{0}+\binom{2003}{1}+\binom{2003}{2}+\cdots+\binom{2003}{2003} \\
\omega^{-2}\left(\omega^{-1}+1\right)^{2003} & =\binom{2003}{0} \omega^{-2}+\binom{2003}{1} \omega^{-3}+\binom{2003}{2} \omega^{-4}+\cdots+\binom{2003}{2003} \omega^{-2005}
\end{aligned}
$$

If we add these together, then the terms $\binom{2003}{n}$ for $n \equiv 1(\bmod 3)$ appear with coefficient 3 , while the remaining terms appear with coefficient $1+\omega+\omega^{2}=0$. Thus the desired sum is just $\left(\omega^{2}(\omega+1)^{2003}+2^{2003}+\omega^{-2}\left(\omega^{-1}+1\right)^{2003}\right) / 3$. Simplifying using $\omega+1=-\omega^{2}$ and $\omega^{-1}+1=-\omega$ gives $\left(-1+2^{2003}+-1\right) / 3=\left(2^{2003}-2\right) / 3$.