Every third

Let $z$ be complex number such that $z^3=1$ .

And

$S={2000 \choose 2} + {2000 \choose 5} + {2000 \choose 8} +..+ {2000 \choose 2000}$

Now take a look at number ( $1+z)^{2000}$

Using the binomial formula and using the definition of $z$ in binomial expand you will have terms with $z$ or $z^2$ or $z^3=1$ . So you can rewrite our number as:

$(1+z)^{2000}={2000 \choose 0} +{2000 \choose 3} +...+{2000 \choose 1998} + z({2000 \choose 1} +{2000 \choose 4} +...+{2000 \choose 1999} )+z^2({2000 \choose 2} + {2000 \choose 5} + {2000 \choose 8} +..+ {2000 \choose 2000})$

Using that ${n\choose k}={n\choose n-k}$ .

${2000 \choose 0} +{2000 \choose 3} +...+{2000 \choose 1998}={2000 \choose 2} + {2000 \choose 5} + {2000 \choose 8} +..+ {2000 \choose 2000}=S$

And because of that the last term can be written as:

${2000 \choose 1} +{2000 \choose 4} +...+{2000 \choose 1999} =2^{2000}-2S$

Now we have:

$(1+z)^{2000}=S+z(2^n-2S)+z^2 S$

Now you can calculate it simple using the fact that if $z^3=1$ than $1+z+z^2=0$

So now

$(1+z)^{2000}=z$

And

$z=S+z(2^n-2S)-S-zS$

$z=z 2^n-z3S$

And finally

$S=\frac{2^{2000}-1}{3}$

Better write the expansion of x(1+x)^2000 .replace x=1,omega,omega^2 in the expanded equation .add the three equations and u get the answer.