If `s_n=(1-4/1)(1-4/9)(1-4/25)......(1-4/((2n-1)^2)),` where `n in N,` then

Options

A. `S_(n) = (1+2n)/(1-2n)`
B. `S_(n) = (1-2n)/(1+2n)`
C. `S_(oo) = - 1`
D. `S_(oo) = - 2`

Answer

To solve the problem, we need to simplify the expression given by \( s_n = (1 - \frac{4}{1})(1 - \frac{4}{9})(1 - \frac{4}{25}) \ldots (1 - \frac{4}{(2n-1)^2}) \).

### Step-by-Step Solution:

1. **Rewrite the Terms**:
Each term in the product can be rewritten using the identity \( 1 - a^2 = (1 - a)(1 + a) \). Here, we can express \( 4 \) as \( 2^2 \):
\[ s_n = \prod_{k=1}^{n} \left(1 - \frac{4}{(2k-1)^2}\right) = \prod_{k=1}^{n} \left(1 - \left(\frac{2}{2k-1}\right)^2\right) \]

2. **Apply the Difference of Squares**:
Using the identity \( a^2 - b^2 = (a - b)(a + b) \), we can rewrite each term:
\[ 1 - \left(\frac{2}{2k-1}\right)^2 = \left(1 - \frac{2}{2k-1}\right)\left(1 + \frac{2}{2k-1}\right) \]
Thus, we have:
\[ s_n = \prod_{k=1}^{n} \left( \left(1 - \frac{2}{2k-1}\right)\left(1 + \frac{2}{2k-1}\right) \right) \]

3. **Simplify Each Factor**:
Now, let's simplify each factor:
- The first factor:
\[ 1 - \frac{2}{2k-1} = \frac{(2k-1) - 2}{2k-1} = \frac{2k-3}{2k-1} \]
- The second factor:
\[ 1 + \frac{2}{2k-1} = \frac{(2k-1) + 2}{2k-1} = \frac{2k+1}{2k-1} \]

4. **Combine the Factors**:
Therefore, we can write:
\[ s_n = \prod_{k=1}^{n} \left(\frac{2k-3}{2k-1} \cdot \frac{2k+1}{2k-1}\right) \]

5. **Simplify the Product**:
The product can be separated:
\[ s_n = \prod_{k=1}^{n} \frac{(2k-3)(2k+1)}{(2k-1)^2} \]
This product will have many terms that cancel out.

6. **Final Expression**:
After simplifying, we find that:
\[ s_n = \frac{(-1)(2n+1)}{(2n-1)} \]
Therefore:
\[ s_n = \frac{1 + 2n}{1 - 2n} \]

### Final Answer:
\[ s_n = \frac{1 + 2n}{1 - 2n} \]