Answer: D

Solution:

Let $z_n = a_n + b_n i.$ Then $$\begin{aligned} z_{n+1} &= (\sqrt{3}\,a_n - b_n) \\ &\quad {}+ (\sqrt{3}\,b_n + a_n)i \\ &= (a_n + b_n i)(\sqrt{3} + i), \end{aligned}$$ so $z_{n+1} = z_n(\sqrt{3} + i)$ and $z_{100} = z_1(\sqrt{3} + i)^{99}.$

Since $\sqrt{3} + i = 2(\cos 30^\circ + i\sin 30^\circ),$ De Moivre's theorem gives $(\sqrt{3} + i)^{99}$ $= 2^{99}(\cos 2970^\circ + i\sin 2970^\circ).$ As $2970^\circ$ is coterminal with $90^\circ,$ this equals $2^{99} i.$

Thus $2 + 4i = z_1 \cdot 2^{99} i,$ so $$z_1 = \dfrac{2 + 4i}{2^{99} i} = \dfrac{4 - 2i}{2^{99}}.$$

Then $a_1 = \tfrac{4}{2^{99}}$ and $b_1 = -\tfrac{2}{2^{99}},$ so $$a_1 + b_1 = \dfrac{2}{2^{99}} = \dfrac{1}{2^{98}}.$$

Thus, D is the correct answer.