If `lta_(n)gtandltb_(n)gt` be 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 `ninN`. Then, `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 \( a_1 a_2 a_3 \ldots a_n \) in terms of the sequence \( b_n \).

### Step-by-Step Solution:

1. **Define the sequences**:
\[ a_n = x^{\frac{1}{2^n}} + y^{\frac{1}{2^n}} \]
\[ b_n = x^{\frac{1}{2^n}} - y^{\frac{1}{2^n}} \]

2. **Multiply and divide \( a_n \)**:
We can rewrite \( a_n \) by multiplying and dividing by \( b_n \):
\[ a_n = \left( x^{\frac{1}{2^n}} + y^{\frac{1}{2^n}} \right) \cdot \frac{b_n}{b_n} \]
This gives us:
\[ a_n = \frac{(x^{\frac{1}{2^n}} + y^{\frac{1}{2^n}})(x^{\frac{1}{2^n}} - y^{\frac{1}{2^n}})}{b_n} \]

3. **Use the difference of squares**:
The numerator can be simplified using the difference of squares:
\[ a_n = \frac{(x^{\frac{1}{2^n}})^2 - (y^{\frac{1}{2^n}})^2}{b_n} \]
This simplifies to:
\[ a_n = \frac{x^{\frac{2}{2^n}} - y^{\frac{2}{2^n}}}{b_n} \]

4. **Express \( a_n \) in terms of \( b_n \)**:
We can express \( a_n \) in terms of \( b_n \) as follows:
\[ a_n = \frac{b_{n-1} b_n}{b_n} \]

5. **Calculate the product \( a_1 a_2 \ldots a_n \)**:
The product can be expressed as:
\[ a_1 a_2 \ldots a_n = \frac{b_0}{b_1} \cdot \frac{b_1}{b_2} \cdots \frac{b_{n-1}}{b_n} \]
Here, all intermediate terms cancel out:
\[ a_1 a_2 \ldots a_n = \frac{b_0}{b_n} \]

6. **Substitute \( b_0 \)**:
Now we substitute \( b_0 \):
\[ b_0 = x^{\frac{1}{2^0}} - y^{\frac{1}{2^0}} = x - y \]
Thus, we have:
\[ a_1 a_2 \ldots a_n = \frac{x - y}{b_n} \]

### Final Result:
The product \( a_1 a_2 a_3 \ldots a_n \) is given by:
\[ a_1 a_2 a_3 \ldots a_n = \frac{x - y}{b_n} \]