Saudi Arabia Mathematical Competitions 2012

The first step gives rise to one black square of area ${\left(\frac{1}{3}\right)}^2=\frac{1}{9}$. After the second step we obtain eight more squares of side $\frac{1}{9}$, the black region increasing thus by $\frac{8}{9^2}$. In the same manner, the third step increases the black area by $8^2=64$ black squares, each of area $\frac{1}{27^2}$, that is at this stage the black area becomes $$\frac{1}{9}+\frac{8}{9^2}+\frac{8^2}{9^3}.$$ We conclude that after $1000$ steps, the area of the black region is $$\begin{array}{rl}\frac{1}{9}+\frac{8}{9^2}+\frac{8^2}{9^3}+\cdots +\frac{8^{999}}{9^{1000}}& =\frac{1}{9}\left(1+\frac{8}{9}+{\left(\frac{8}{9}\right)}^2+\cdots +{\left(\frac{8}{9}\right)}^{999}\right)\\ & =\frac{1}{9}\cdot \frac{1-{\left(\frac{8}{9}\right)}^{1000}}{1-\frac{8}{9}}=1-{\left(\frac{8}{9}\right)}^{1000}.\end{array}$$ It remains to prove that the last number is greater than $0.999$, i.e., $1-{\left(\frac{8}{9}\right)}^{1000}>0.999$, or equivalently ${\left(\frac{8}{9}\right)}^{1000}<0.001$.

This easily follows by using a binomial expansion: $${\left(\frac{9}{8}\right)}^{1000}={\left(1+\frac{1}{8}\right)}^{1000}>\left(\frac{1000}{2}\right)\cdot \frac{1}{8^2}=\frac{1000\cdot 999}{2\cdot 64}>1000.$$ This completes the proof.