Am I incorrect in the assumption that I deal with $\sqrt{3}$ in the same way I would approach $\sqrt{2}$, by adding and subtracting $1$ such that:

$\sqrt{3}=1+\sqrt{3}-1=1+\cfrac{2}{1+\sqrt{3}}$

The table representation is $[1; 1,2]$.

This is what I have so far:

$\sqrt{3}=1+\cfrac{2}{2+\cfrac{2}{1+\sqrt{3}}}$

Which seems to lead me to $[1;2,2,...]$ any help would be appreciated, thanks