Let `C_(0)` be circle of radius 1. For `nge1`, let `C_(n)` be a circle whose area equals the area of a squre inscibed in `C_(n-1)`. Then `sum underset(i-0)(oo)` Area `(C_(i))` equals

Options

A. `pi^(2)`
B. `(pi-2)/(pi^(2))`
C. `(1)/(pi^(2))`
D. `(pi^(2))/(pi-2)`

Answer

D

Explanation

`sum_(i-)^(oo)` Area `(C_(i))=pir_(0)^(2)+pir_(1)^(2)+pi_(2)^(2)+pir_(3)^(2)+....oo`

Area of `C_(n)=pir_(n)^(2)=(sqrt(2)r_(n-1)^(2))^(2)`
`r_(n)^(2)=(2)/(pi)r_(n-1)^(2)`
so `r_(1)^(2)=(2)/(pi)r_(0)^(2),r_(2)^(2)=(2)/(pi)r_(1)^(2)`
`=(2)/(pi)((2)/(pi)r_(0)^(2))`
`r_(3)^(2)=(2)/(pi)(r_(2)^(2))=(2)/(pi)((2)/(pi)(2)/(pi)r_(0)^(2))`
`So, sum_(i=o)^(oo)` Area `(C_i)=pi[r_(0)^(2)+(2)/(pi)r_(0)^(2)+(2)/(pi)*(2)/(pi)r_(0)^(2)+....oo]`
`=(pir_(0)^(2))/(1-(2)/(pi))=(pi^(2)r_(0)^(2))/(pi-2)AAr_(0)=1`
`=(pi^(2))/(pi-2)`