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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 $\mathrm{C}_{\mathrm{n}-1} .$ Then $\sum_{\mathrm{i}=0}^{\infty}$ Area $\left(\mathrm{C}_{\mathrm{i}}\right)$ equals:
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Verified Answer
The correct answer is:
$\frac{\pi^{2}}{\pi-2}$
Let $a, a_{1}, a_{2--}$ be the length of siden of square inscribed in circle $c_{0}, c_{1}, c_{2}=$ and $\mathrm{r}_{1}, \mathrm{r}_{2}, \mathrm{r}_{3}, \ldots_{----}$be radius of circle $\mathrm{c}_{1}, \mathrm{c}_{2}$, then, $a_{0}^{2}=2=\pi r_{1}^{2}$
$$
\begin{array}{l}
r_{1}=\sqrt{\frac{2}{\pi}} \\
2 a_{1}^{2}=\frac{8}{\pi} \quad a_{1}^{2}=\frac{4}{\pi}=\pi r_{2}^{2} \quad r_{2}=\frac{2}{\pi} \\
\sum_{i=0}^{\infty} \text { Area }(C i)=\pi . i \\
=\pi+2+\frac{4}{\pi}+\frac{8}{\pi^{2}}=\frac{\pi}{1-\frac{2}{\pi}}=\frac{\pi^{2}}{\pi-2}
\end{array}
$$
$$
\begin{array}{l}
r_{1}=\sqrt{\frac{2}{\pi}} \\
2 a_{1}^{2}=\frac{8}{\pi} \quad a_{1}^{2}=\frac{4}{\pi}=\pi r_{2}^{2} \quad r_{2}=\frac{2}{\pi} \\
\sum_{i=0}^{\infty} \text { Area }(C i)=\pi . i \\
=\pi+2+\frac{4}{\pi}+\frac{8}{\pi^{2}}=\frac{\pi}{1-\frac{2}{\pi}}=\frac{\pi^{2}}{\pi-2}
\end{array}
$$
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