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 \( a_1 a_2 a_3 \ldots a_n \) given the sequences:

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

### Step 1: Calculate the product \( a_n \cdot b_n \)

We start by multiplying \( a_n \) and \( b_n \):

\[ a_n \cdot b_n = (x^{\frac{1}{2^n}} + y^{\frac{1}{2^n}})(x^{\frac{1}{2^n}} - y^{\frac{1}{2^n}}) \]

Using the difference of squares formula, we have:

\[ a_n \cdot b_n = (x^{\frac{1}{2^n}})^2 - (y^{\frac{1}{2^n}})^2 \]

This simplifies to:

\[ a_n \cdot b_n = x^{\frac{2}{2^n}} - y^{\frac{2}{2^n}} = x^{\frac{1}{2^{n-1}}} - y^{\frac{1}{2^{n-1}}} \]

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

From the previous step, we can express \( a_n \) in terms of \( b_n \):

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

### Step 3: Calculate the product \( a_1 a_2 a_3 \ldots a_n \)

Now, we can express the product \( a_1 a_2 a_3 \ldots a_n \):

\[ a_1 a_2 a_3 \ldots a_n = \prod_{k=1}^{n} a_k \]

Using the relationship we derived:

\[ a_k = \frac{b_k}{b_{k-1}} \quad \text{(for } k \geq 2 \text{)} \]

Thus, we can write:

\[ a_1 a_2 a_3 \ldots a_n = a_1 \cdot \frac{b_2}{b_1} \cdot \frac{b_3}{b_2} \cdots \frac{b_n}{b_{n-1}} \]

Notice that all terms cancel except for \( a_1 \) and \( b_n \):

\[ a_1 a_2 a_3 \ldots a_n = a_1 \cdot \frac{b_n}{b_1} \]

### Step 4: Calculate \( a_1 \) and \( b_1 \)

Now we need to find \( a_1 \) and \( b_1 \):

\[ a_1 = x^{\frac{1}{2^1}} + y^{\frac{1}{2^1}} = x^{\frac{1}{2}} + y^{\frac{1}{2}} \]
\[ b_1 = x^{\frac{1}{2^1}} - y^{\frac{1}{2^1}} = x^{\frac{1}{2}} - y^{\frac{1}{2}} \]

### Step 5: Final expression

Putting it all together, we get:

\[ a_1 a_2 a_3 \ldots a_n = \frac{(x^{\frac{1}{2}} + y^{\frac{1}{2}}) \cdot b_n}{b_1} \]

### Conclusion

Thus, the final expression for \( a_1 a_2 a_3 \ldots a_n \) is:

\[ \boxed{\frac{(x^{\frac{1}{2}} + y^{\frac{1}{2}}) \cdot (x^{\frac{1}{2^{n-1}}} - y^{\frac{1}{2^{n-1}}})}{(x^{\frac{1}{2}} - y^{\frac{1}{2}})}} \]