To solve this problem, we need to find the correct series expansion for the function:

$f(x) = \frac{1 - x^2}{1 + x^2}.$

Step-by-step Solution:

1. Rewrite the function:

   Notice that the function can be rewritten in a more manageable form: $f(x) = \frac{1 - x^2}{1 + x^2} = \frac{1 + x^2 - 2x^2}{1 + x^2} = 1 - \frac{2x^2}{1 + x^2}.$

   So, we can express $f(x)$ as: $f(x) = 1 - 2 \cdot \frac{x^2}{1 + x^2}.$

2. Find the series expansion for $\frac{1}{1 + x^2}$:

   We recognize that $\frac{1}{1 + x^2}$ is a geometric series for $|x^2| < 1$: $\frac{1}{1 + x^2} = \sum_{n=0}^{\infty} (-1)^n x^{2n}.$

3. Substitute the series into $f(x)$:

   Now, substitute the series expansion into the expression for $f(x)$:

   $f(x) = 1 - 2 \cdot \left(x^2 \sum_{n=0}^{\infty} (-1)^n x^{2n}\right).$

   This simplifies to:

   $f(x) = 1 - 2 \sum_{n=0}^{\infty} (-1)^n x^{2n+2}.$

4. Adjust the index for the series:

   To match the series to the options given, we change the index in the series by substituting $m = n+1$:

   $f(x) = 1 - 2 \sum_{m=1}^{\infty} (-1)^{m-1} x^{2m}.$

   Simplify the exponent signs:

   $f(x) = 1 - 2 \sum_{n=0}^{\infty} (-1)^n x^{2n}.$

5. Combine the constant term:

   The series starts from $n = 0$, so it becomes:

   $f(x) = -1 + 2 \sum_{n=0}^{\infty} (-1)^n x^{2n}.$

Conclusion:

This matches with Option A:

$\textbf{A. } -1 + 2 \sum_{n=0}^{\infty} (-1)^n x^{2n}.$

Final Answer:

The correct answer is (a).

Would you like further details or explanations? Here are 5 related questions you might find interesting:

1. How do you derive the geometric series formula?
2. What are the convergence conditions for a series expansion?
3. How do you change the index of summation in a series?
4. Can you derive Taylor series for more complex functions?
5. What is the radius of convergence for a series like $\sum (-1)^n x^{2n}$?

Tip: When working with series, always check the radius of convergence to ensure the series representation is valid.