If `A_1` is the area area bounded by `|x-a_i| + |y|=b_i, i in N` ,where `a_(i+1) =a_i+3/2 b_i and b_(i+1)=b_i/2,a_i=0 and b_i=32`, then

Options

A. `A_(3)=128`
B. `A_(3)=256`
C. `underset(nrarroo)limoverset(n)underset(i=1)SigmaA_(i)=(8)/(3)(32)^(2)`
D. `underset(nrarroo)limoverset(n)underset(i=1)SigmaA_(i)=(4)/(3)(16)^(2)`

Explanation

To solve the given problem step by step, we will first analyze the area bounded by the equation \( |x - a_i| + |y| = b_i \) and then find the required areas \( a_i \) for \( i = 1, 2, 3 \).

### Step 1: Understanding the Area

The equation \( |x - a_i| + |y| = b_i \) represents a diamond (or rhombus) shape in the coordinate plane. The vertices of this diamond are located at:
- \( (a_i + b_i, 0) \)
- \( (a_i - b_i, 0) \)
- \( (a_i, b_i) \)
- \( (a_i, -b_i) \)

The area \( A \) of a diamond can be calculated using the formula:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
where \( d_1 \) and \( d_2 \) are the lengths of the diagonals. In this case, the diagonals are \( 2b_i \) and \( 2b_i \), so:
\[ A = \frac{1}{2} \times (2b_i) \times (2b_i) = 2b_i^2 \]

### Step 2: Calculate \( a_1 \)

Given:
- \( a_1 = 0 \)
- \( b_1 = 32 \)

The area \( a_1 \) is:
\[ a_1 = 2b_1^2 = 2 \times (32)^2 = 2 \times 1024 = 2048 \]

### Step 3: Calculate \( a_2 \) and \( b_2 \)

Using the recurrence relations:
- \( a_{i+1} = a_i + \frac{3}{2} b_i \)
- \( b_{i+1} = \frac{b_i}{2} \)

For \( i = 1 \):
\[ a_2 = a_1 + \frac{3}{2} b_1 = 0 + \frac{3}{2} \times 32 = 48 \]
\[ b_2 = \frac{b_1}{2} = \frac{32}{2} = 16 \]

Now, calculate \( a_2 \):
\[ a_2 = 2b_2^2 = 2 \times (16)^2 = 2 \times 256 = 512 \]

### Step 4: Calculate \( a_3 \) and \( b_3 \)

For \( i = 2 \):
\[ a_3 = a_2 + \frac{3}{2} b_2 = 48 + \frac{3}{2} \times 16 = 48 + 24 = 72 \]
\[ b_3 = \frac{b_2}{2} = \frac{16}{2} = 8 \]

Now, calculate \( a_3 \):
\[ a_3 = 2b_3^2 = 2 \times (8)^2 = 2 \times 64 = 128 \]

### Step 5: Sum of Areas as \( n \to \infty \)

The areas \( a_i \) form a geometric progression:
- \( a_1 = 2048 \)
- \( a_2 = 512 \)
- \( a_3 = 128 \)

The common ratio \( r \) can be calculated as:
\[ r = \frac{a_2}{a_1} = \frac{512}{2048} = \frac{1}{4} \]

The sum of the infinite geometric series is given by:
\[ S = \frac{a_1}{1 - r} = \frac{2048}{1 - \frac{1}{4}} = \frac{2048}{\frac{3}{4}} = 2048 \times \frac{4}{3} = \frac{8192}{3} \]

### Final Result

Thus, we have:
- \( a_3 = 128 \)
- The limit as \( n \to \infty \) of the sum of areas \( \sum_{i=1}^{n} a_i = \frac{8192}{3} \)