Let $$S_1$$ be a square of side 4 cm. Circle $$C_1$$ circumscribes the square $$S_1$$ such that all its corners are on $$C_1$$. Another square $$S_2$$ circumscribes the circle $$C_1$$. Circle $$C_2$$ circumscribes the square $$S_2$$, and square $$S_3$$ circumscribes circle $$C_2$$, & so on. If $$A_N$$ is the area between the square $$S_N$$ and the circle $$C_N$$, where N is the natural number. then the ratio of sum of all $$A_N$$ to $$A_l$$ is

 Let the side of the square x be denoted by $$S_x$$

Let the radius of the circle y be denoted by $$r_y$$ and diameter by $$d_y$$

$$S_{s1}=4$$

$$r_{c1}=\frac{d_{c1}}{2}=\frac{4\sqrt{\ 2}}{2}=2\sqrt{\ 2}$$

$$S_{s2}=d_{c1}=4\sqrt{\ 2}$$

$$r_{c2}=4$$

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A1 = $$\pi\ r_{c1}^2-S_{s1}^2\ =\ 8\pi\ -16$$

A2 = $$16\pi\ -32$$

A3 = $$32\pi\ -64$$

Sum = $$\frac{\left(8\pi\ -16\right)\left(2^n-1\right)}{2-1}\ =\ \left(8\pi\ -16\right)\left(2^n-1\right)$$

Ratio = $$\left(2^n-1\right)$$

Since the value of n is unknown, the answer can be C or D