1. As shown in Figure 1, the side length of the regular hexagon $A_{1} B_{1} C_{1} D_{1} E_{1} F_{1}$ is 1, and its six diagonals form another regular hexagon $A_{2} B_{2} C_{2} D_{2} E_{2} F_{2}$. This process continues. Then the sum of the areas of all these hexagons is $\qquad$
-1. $\frac{9 \sqrt{3}}{4}$.
Let the area of the $n$-th regular hexagon be $a_{n}$. Then $a_{1}=6 \times \frac{\sqrt{3}}{4} \times 1^{2}=\frac{3 \sqrt{3}}{2}$.

It is easy to see that $A_{2} B_{2}=A_{2} B_{1}=\frac{\sqrt{3}}{3}, \frac{a_{2}}{a_{1}}=\frac{1}{3}$.
In general, $\frac{a_{n+1}}{a_{n}}=\frac{1}{3}$.
Therefore, $\sum_{i=1}^{\infty} a_{i}=\frac{a_{1}}{1-\frac{1}{3}}=\frac{3}{2} a_{1}=\frac{9 \sqrt{3}}{4}$.