${2000\choose1}+{2000\choose4}+{2000\choose7}+\cdots +{2000\choose1996}+{2000\choose1999}=?$ ${2000\choose1}+{2000\choose4}+{2000\choose7}+\cdots +{2000\choose1996}+{2000\choose1999}=?$
 A: $$(1+x)^n=\sum_{r=0}^n\binom nr x^r$$ where $3\nmid n$
Set $x=1,w,w^2$ where $w$ is a complex cube root of unity
$$P=(1+1)^n=\sum_{r=0}^n\binom nr 1^r$$
$$Q=(1+w)^n=\sum_{r=0}^n\binom nr w^r$$
$$R=(1+w^2)^n=\sum_{r=0}^n\binom nr w^{2r}$$
Find $P+w^2\cdot Q+w\cdot R$
Use $1+w+w^2=0$ to simplify the calculation 
A: a = ${2000\choose1}+{2000\choose4}+{2000\choose7}+\cdots +{2000\choose1996}+{2000\choose1999}$
b = ${2000\choose0}+{2000\choose3}+{2000\choose6}+\cdots +{2000\choose1995}+{2000\choose1998} = {2000\choose2}+{2000\choose5}+{2000\choose8}+\cdots +{2000\choose1997}+{2000\choose2000}$
and that $b = \dfrac{(1+1)^{2000} + (1+\omega)^{2000} + (1+\omega^2)^{2000}}{3}$. where $\omega$ is the primitive third root of unity.  It can be seen that $1+\omega$ is the primitive sixth root of unity and $(1+\omega^2)$ is its inverse
So $(1+\omega)^{2000} = (1+\omega)^2 = \omega$
Similarly $(1+\omega^2)^{2000} = \omega^2$ 
Therefore, $b = \frac{2^{2000}-1}{3}$
$a+2b = 2^{2000}$
$a = 2^{2000} - 2b$
$a = 2^{2000} - 2\frac{2^{2000}-1}{3} = \frac{2^{2000}+2}{3} = \frac{2^{2000}+2}{3}$
