Let C_0 be a circle of radius 1. For n geq 1, | vedclass

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(D) Let $r_n$ be the radius of circle $C_n$. Given $r_0 = 1$,so $	ext{Area}(C_0) = \pi r_0^2 = \pi$.$A$ square inscribed in a circle of radius $r$ ha

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$\frac{\pi^2}{\pi-2}$

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(D) Let $r_n$ be the radius of circle $C_n$. Given $r_0 = 1$,so $	ext{Area}(C_0) = \pi r_0^2 = \pi$.$A$ square inscribed in a circle of radius $r$ has a diagonal equal to the diameter $2r$. If the side of the square is $a$,then $a^2 + a^2 = (2r)^2$,which implies $2a^2 = 4r^2$,so $a^2 = 2r^2$.The area of the square inscribed in $C_{n-1}$ is $2r_{n-1}^2$. By definition,$	ext{Area}(C_n) = \pi r_n^2 = 2r_{n-1}^2$.Thus,$r_n^2 = \frac{2}{\pi} r_{n-1}^2$. This is a geometric progression for the areas $A_n = 	ext{Area}(C_n)$ with first term $A_0 = \pi$ and common ratio $k = \frac{2}{\pi}$.The sum of the areas is $\sum_{i=0}^{\infty} A_i = A_0 + A_0 k + A_0 k^2 + \dots = \frac{A_0}{1-k}$.Substituting the values: $\sum_{i=0}^{\infty} 	ext{Area}(C_i) = \frac{\pi}{1 - \frac{2}{\pi}} = \frac{\pi}{\frac{\pi-2}{\pi}} = \frac{\pi^2}{\pi-2}$.

Let $C_0$ be a circle of radius $1$. For $n \geq 1$,let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1}$. Then,$\sum_{i=0}^{\infty} \text{Area}(C_i)$ equals

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