Sum of Binomial Series of form $\binom{2000}{3k-1}$

Question

Find the Value of $$ \binom{2000}{2}+\binom{2000}{5}+\binom{2000}{8}+\cdots+\binom{2000}{1997}+\binom{2000}{2000}$$

Answer 1

Let $$ X=\binom{2000}{2}+\binom{2000}{5}+\binom{2000}{8}+\cdots+\binom{2000}{1997}+\binom{2000}{2000}$$ $X$ can also be written as:

$$ X=\binom{2000}{0}+\binom{2000}{3}+\binom{2000}{6}+\cdots+\binom{2000}{1995}+\binom{2000}{1998} $$

Let $$ Y= \binom{2000}{1}+\binom{2000}{4}+\binom{2000}{7}+\cdots+\binom{2000}{1996}+\binom{2000}{1999}$$ Adding all:

$$ 2X+Y=2^{2000}$$

I need small hint to proceed further

Answer 2

You're off to a good start! In your post you defined quantities $X$ and $Y$ and found that $2X+Y=2^{2000}$. This post builds on that by giving you a second relation between $X$ and $Y$.

Let $\omega=e^{2\pi i/3}$. So $\omega^3=1$. Then by the binomial theorem, $$(1+\omega)^{2000}=\binom{2000}{0}+\binom{2000}{1}\omega+\binom{2000}{2}\omega^2+\ldots+\binom{2000}{2000}\omega^2$$

Now we compare the real parts of each side. $$\Re((1+\omega)^{2000})=\Re(\binom{2000}{0}+\binom{2000}{1}\omega+\binom{2000}{2}\omega^2+\ldots+\binom{2000}{2000}\omega^2)$$

$$1+\omega=\frac{1+\sqrt3 i}{2}=e^{\pi i/3}$$ so $(1+\omega)^{2000}=\omega$, and we know that $\Re (\omega)=\Re (\omega^2)=\frac{-1}{2}$. Plugging in above, we see that

$$\frac{-1}{2}=\binom{2000}{0}-\frac{1}{2}\binom{2000}{1}-\frac{1}{2}\binom{2000}{2}+\binom{2000}{3}+\ldots-\frac{1}{2}\binom{2000}{2000}$$

This gives us a second equation, $\frac{-1}{2}=X-\frac{Y}{2}-\frac{X}{2}$.

Now you have 2 equations in 2 unknowns, and you just need to solve for $X$.