## Algebra 2 (1st Edition)

$b_n= (\dfrac{8}{9})^n S\\b_{12}= (\dfrac{8}{9})^{12} S$
Let $S$ be the area of the initial square. Here, we have $b_1= S-\dfrac{S}{9}$ and $b_2= \dfrac{8S}{9}-(8) \dfrac{S}{(9)^2}=\dfrac{64S}{81}$ Likewise, we have $b_n= (\dfrac{8}{9})^n S$ Hence, $b_{12}= (\dfrac{8}{9})^{12} S$