smc

Let $(x,y)=(a_1,b_1)$. Then, we can begin to list out terms as follows: $(a_2,b_2)=(x\sqrt{3}-y,y\sqrt{3}+x)$ $(a_3,b_3)=(2x-2y\sqrt{3},2y+2x\sqrt{3})$ $(a_4,b_4)=(-8y,8x)$ We notice that the sequence follows the rule $(a_{n+3},b_{n+3})=(-2^3{b}_n,2^3{a}_n)$ We can now start listing out every third point, getting: $(a_1,b_1)=(x,y)$ $(a_4,b_4)=(-2^{3}y,2^{3}x)$ $(a_7,b_7)=(-2^{6}x,-2^{6}y)$ $(a_{10},b_{10})=(2^{9}y,-2^{9}x)$ $(a_{13},b_{13})=(2^{12}x,2^{12}y)$ We can make two observations from this: (1) In $a_n$, the coefficient of $x$ and $y$ is $2^{n-1}$ (2) The positioning of $x$ and $y$, and their signs, cycle with every $12$ terms. We know then that from (1), the coefficients of $x$ and $y$ in $(a_{100},b_{100})$ are both $2^{99}$ We can apply (2), finding $100\text{(mod }12)=4$, so the positions and signs of $x$ and $y$ are the same in $(a_{100},b_{100})$ as they are in $(a_4,b_4)$. From this, we can get $(a_{100},b_{100})=(-2^{99}y,2^{99}x)$. We know that $(a_{100},b_{100})=(2,4)$, so we get the following: $-2^{99}y=2\Rightarrow y=-\frac{1}{2^{98}}$ $2^{99}x=4\Rightarrow x=\frac{1}{2^{97}}$ The answer is $x+y=\frac{1}{2^{97}}-\frac{1}{2^{98}}=\boxed{\frac{1}{2^{98}}}$..