$ABCDEFGH$ , which shall be named $O_1$ , has squares $ACEG$ and $BDFH$ inscribed in it. These squares form a smaller octagon, $O_2$ .

A regular octagon$O_2$ has squares inscribed in it in the same manner, and the resulting hexagon formed is named $O_3$ . This pattern is repeated until infinity.

Let the area of octagon $O_i$ be denoted by $A_i$ . Given that $A_1=1$ , for some integers $a$ and $b$ , where $a$ is square-free, $\large \sum _{ i=1 }^{ \infty }{ { A }_{ i } } =\sqrt { a } +b$ Find $a+b$ .

The answer is 3.

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Using the hint, we know that the geometric progression $A_1,A_2,A_3,...$ has the ratio $r= 2-\sqrt2$ Therefore, the sum $S=\frac{1}{1-(2-\sqrt2)} =\sqrt2+1$

Hence $a=2 , b=1$ and $a+b=3$