Calculating $\binom{99}{0}+ \binom{100}{2}+ \binom{99}{3}+ \binom{100}{5}+ \binom{99}{6}+ \binom{100}{8}+ ..+ \binom{100}{98}+\binom{99}{99}$

Question

Hello everyone how can I calculate:

$\binom{99}{0}+ \binom{100}{2}+ \binom{99}{3}+ \binom{100}{5}+ \binom{99}{6}+ \binom{100}{8}+ \cdots+ \binom{100}{98}+\binom{99}{99}$ ?

I tried to mark $\omega = \frac{1}{2}-\frac{\sqrt{3}i}{2}$ and calculate $\binom{99}{0}+ \binom{99}{3}+\binom{99}{6}+ \cdots+\binom{99}{99}$ by $(1+1)^{99} +(1+\omega)^{99} + (1+\overline\omega)^{99} =\frac{2^{99} +1}{3}$

But I don't know how to continue.

Answer 1

Hint

$$\binom{100}{k}=\binom{99}{k}+\binom{99}{k-1}$$