`{a_n} and {b_n}` are two sequences given by `a_n=(x)^(1//2^n) +(y)^(1//2^n) and b_(n)=(x)^(1//2^n)-(y)^(1//2^n)` for all n `in` N. The value of `a_1 a_2 a_3……..a_n` is equal to

Options

A. `x-y`
B. `(x+y)/(b_n)`
C. `(x-y)/(b_n)`
D. `(xy)/(b_n)`

Answer

To solve the problem, we need to find the product \( a_1 a_2 a_3 \ldots a_n \) where the sequences \( a_n \) and \( b_n \) are defined as follows:

\[ a_n = x^{1/2^n} + y^{1/2^n} \]
\[ b_n = x^{1/2^n} - y^{1/2^n} \]

### Step 1: Write down the product \( a_1 a_2 a_3 \ldots a_n \)

We start by calculating the product of the first \( n \) terms of the sequence \( a_n \):

\[ P = a_1 a_2 a_3 \ldots a_n = (x^{1/2} + y^{1/2})(x^{1/4} + y^{1/4})(x^{1/8} + y^{1/8}) \ldots (x^{1/2^n} + y^{1/2^n}) \]

### Step 2: Express \( P \) in terms of \( b_n \)

Next, we can relate this product to the sequence \( b_n \). We know that:

\[ a_n = (x^{1/2^n} + y^{1/2^n}) \quad \text{and} \quad b_n = (x^{1/2^n} - y^{1/2^n}) \]

We can use the identity \( (a + b)(a - b) = a^2 - b^2 \) to express the product:

\[ P \cdot b_1 b_2 b_3 \ldots b_n = (x^{1/2} + y^{1/2})(x^{1/2} - y^{1/2})(x^{1/4} + y^{1/4})(x^{1/4} - y^{1/4}) \ldots (x^{1/2^n} + y^{1/2^n})(x^{1/2^n} - y^{1/2^n}) \]

### Step 3: Simplify the expression

Using the above identity, we can simplify the product:

\[ P \cdot b_1 b_2 b_3 \ldots b_n = (x^{1/2})^2 - (y^{1/2})^2 + (x^{1/4})^2 - (y^{1/4})^2 + \ldots + (x^{1/2^n})^2 - (y^{1/2^n})^2 \]

This results in:

\[ = x^{1} - y^{1} = x - y \]

### Step 4: Find the product of \( b_n \)

Next, we need to compute \( b_1 b_2 b_3 \ldots b_n \):

\[ b_n = x^{1/2^n} - y^{1/2^n} \]

Thus, the product \( b_1 b_2 b_3 \ldots b_n \) can be calculated as:

\[ b_1 b_2 b_3 \ldots b_n = (x^{1/2} - y^{1/2})(x^{1/4} - y^{1/4})(x^{1/8} - y^{1/8}) \ldots (x^{1/2^n} - y^{1/2^n}) \]

### Step 5: Combine results

Now we can express \( P \):

\[ P = \frac{x - y}{b_1 b_2 b_3 \ldots b_n} \]

### Final Result

Thus, the value of \( a_1 a_2 a_3 \ldots a_n \) is given by:

\[ a_1 a_2 a_3 \ldots a_n = \frac{x - y}{b_1 b_2 b_3 \ldots b_n} \]