If `x_(n)=cos(pi/2^(n))+isin(pi/2^(n)), n in N` then `x_(1),x_(2),x_(3)………………..x_(infty)`. Is equal to

Options

A. 1
B. `-1`
C. 0
D. none of these

Explanation

To solve the problem, we need to find the product \( x_1 \times x_2 \times x_3 \times \ldots \times x_\infty \) where \( x_n = \cos\left(\frac{\pi}{2^n}\right) + i \sin\left(\frac{\pi}{2^n}\right) \).

### Step-by-Step Solution:

1. **Express \( x_n \) in Exponential Form**:
\[ x_n = \cos\left(\frac{\pi}{2^n}\right) + i \sin\left(\frac{\pi}{2^n}\right) = e^{i \frac{\pi}{2^n}} \]

2. **Write the Product**:
We need to find the product:
\[ x_1 \times x_2 \times x_3 \times \ldots = e^{i \frac{\pi}{2^1}} \times e^{i \frac{\pi}{2^2}} \times e^{i \frac{\pi}{2^3}} \times \ldots \]

3. **Combine the Exponents**:
Using the property of exponents, we can combine the product:
\[ x_1 \times x_2 \times x_3 \times \ldots = e^{i \left(\frac{\pi}{2^1} + \frac{\pi}{2^2} + \frac{\pi}{2^3} + \ldots\right)} \]

4. **Identify the Series**:
The series inside the exponent is:
\[ \frac{\pi}{2} + \frac{\pi}{4} + \frac{\pi}{8} + \ldots \]
This is a geometric series where the first term \( a = \frac{\pi}{2} \) and the common ratio \( r = \frac{1}{2} \).

5. **Sum the Infinite Geometric Series**:
The sum \( S \) of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
Substituting the values:
\[ S = \frac{\frac{\pi}{2}}{1 - \frac{1}{2}} = \frac{\frac{\pi}{2}}{\frac{1}{2}} = \pi \]

6. **Substitute Back to the Exponential**:
Now substitute back into the exponent:
\[ x_1 \times x_2 \times x_3 \times \ldots = e^{i \pi} \]

7. **Evaluate \( e^{i \pi} \)**:
Using Euler's formula:
\[ e^{i \pi} = \cos(\pi) + i \sin(\pi) = -1 + 0i = -1 \]

### Final Result:
Thus, the product \( x_1 \times x_2 \times x_3 \times \ldots \) converges to:
\[ \boxed{-1} \]