Question

Let $C_0$ be a circle of radius $1$. for $n\ge 1$, let $C_n$ be a circle whose area equals the area of a square inscribed in $C_{n-1}$. Then ${\sum }_{i=0}^{\infty }$. Area $\left(C_i\right)$ equals.

• A. ${\pi }^2$
• B. $\frac{\pi -2}{{\pi }^2}$
• C. $\frac{1}{{\pi }^2}$
• D. $\frac{{\pi }^2}{\pi -2}$

Answer 1

The correct option is D $\frac{{\pi }^2}{\pi -2}$

Area of $C_0=\pi$

$C_1=$ Circle having area $2$ since the square inscribed in $C_0$ has area $2$

For a circle of area 2, the radius would be $\sqrt{\frac{2}{\pi }}$

Corresponding area of the square inscribed in it would be $2\times \frac{2}{\pi }$

$\therefore C_2=\frac{4}{\pi }$

$\sum _{i=1}^{\infty }Area\left(C_i\right)=\pi +2+\frac{4}{\pi }+...$

$=\frac{\pi }{1-\frac{2}{\pi }}$

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