There is an infinite sequence of squares such that vertices of every next square lie in the center of the sides of the previous square. Odd-numbered squares are filled with white color, and even-numbered are filled with black. Find the black area as a fraction of the first square.
Suggestions : 1/3 or 1/2 or 2/3 or 3/4
Please help.
If the side length of a square is $\ell$, then the side length of the inner square whose vertices are the centers of the original square is $\dfrac{\ell}{\sqrt{2}}$, and thus its area is $\dfrac{\ell^2}{2}$, which represents half of the area of the original square. Since the first square is colored in white, then the number we are looking for is : $1-\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\cdots\right)=1-\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{2^n}$
It is well known that $\displaystyle\sum_{n=0}^{+\infty}\dfrac{(-1)^n}{2^n}=\dfrac{2}{3}$ and therefore the answer is $\dfrac{1}{3}$