A square is inscribed in circle of radius R, a circle is inscribed in the square, a new square in the circle and so on for n times.
Sum of the areas of all circles is

Options

A. `4piR^(2)(1-(1/2)^(n))`
B. `2piR^(2)(1-(1/2)^(n))`
C. `3piR^(2)(1-(1/3)^(n))`
D. `piR^(2)(1-(1/2)^(n))`

Answer

To find the sum of the areas of all circles inscribed in squares and circles for \( n \) iterations, we can follow these steps:

### Step 1: Understand the Problem
We start with a circle of radius \( R \). A square is inscribed in this circle, and then a circle is inscribed in that square. This process continues for \( n \) iterations.

### Step 2: Determine the Radius of the Circles
1. **First Circle (Outer Circle)**: The area of the first circle is given by:
\[ A_1 = \pi R^2 \]

2. **First Inscribed Square**: The side length of the square inscribed in the circle is \( R\sqrt{2} \) (since the diagonal of the square equals the diameter of the circle). The radius of the circle inscribed in this square is:
\[ r_1 = \frac{R}{\sqrt{2}} \]

3. **Second Circle**: The area of the second circle is:
\[ A_2 = \pi \left(\frac{R}{\sqrt{2}}\right)^2 = \pi \frac{R^2}{2} \]

4. **Second Inscribed Square**: The radius of the next circle inscribed in the square from the previous step is:
\[ r_2 = \frac{r_1}{\sqrt{2}} = \frac{R}{2} \]

5. **Third Circle**: The area of the third circle is:
\[ A_3 = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} \]

### Step 3: Generalize the Radius and Area
Continuing this process, we can see a pattern:
- The radius of the \( n \)-th circle can be expressed as:
\[ r_n = \frac{R}{(\sqrt{2})^{n-1}} = \frac{R}{2^{(n-1)/2}} \]
- The area of the \( n \)-th circle is:
\[ A_n = \pi \left(\frac{R}{2^{(n-1)/2}}\right)^2 = \pi \frac{R^2}{2^{n-1}} \]

### Step 4: Sum the Areas of All Circles
The total area of all circles up to \( n \) is:
\[ S_n = A_1 + A_2 + A_3 + \ldots + A_n = \pi R^2 \left(1 + \frac{1}{2} + \frac{1}{4} + \ldots + \frac{1}{2^{n-1}}\right) \]

This series is a geometric series with:
- First term \( a = 1 \)
- Common ratio \( r = \frac{1}{2} \)

The sum of the first \( n \) terms of a geometric series is given by:
\[ S_n = a \frac{1 - r^n}{1 - r} = 1 \cdot \frac{1 - \left(\frac{1}{2}\right)^n}{1 - \frac{1}{2}} = 2 \left(1 - \frac{1}{2^n}\right) \]

### Step 5: Final Expression for the Total Area
Thus, the total area of all circles is:
\[ S_n = \pi R^2 \cdot 2 \left(1 - \frac{1}{2^n}\right) = 2\pi R^2 \left(1 - \frac{1}{2^n}\right) \]

### Final Answer
The sum of the areas of all circles after \( n \) iterations is:
\[ \boxed{2\pi R^2 \left(1 - \frac{1}{2^n}\right)} \]