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The problem asks to expand the expression \(\left(\frac{1 - x}{1 + x}\right)^n\) in ascending powers of \(x\), up to and including the term in \(x^2\).

### Step-by-step solution:

1. **Rewrite the expression**:  
   \[
   \left(\frac{1 - x}{1 + x}\right)^n = (1 - x)^n (1 + x)^{-n}
   \]
   This gives us two binomial expansions, one for \((1 - x)^n\) and one for \((1 + x)^{-n}\).

2. **Expand \((1 - x)^n\)** using the binomial theorem:
   \[
   (1 - x)^n = 1 - nx + \frac{n(n-1)}{2}x^2 + \cdots
   \]
   This gives us terms up to \(x^2\):
   \[
   (1 - x)^n = 1 - nx + \frac{n(n-1)}{2}x^2
   \]

3. **Expand \((1 + x)^{-n}\)** using the binomial theorem:
   \[
   (1 + x)^{-n} = 1 - nx + \frac{n(n+1)}{2}x^2 + \cdots
   \]
   This gives us terms up to \(x^2\):
   \[
   (1 + x)^{-n} = 1 - nx + \frac{n(n+1)}{2}x^2
   \]

4. **Multiply the expansions**: Now, multiply the two expansions up to the term in \(x^2\).

   \[
   (1 - nx + \frac{n(n-1)}{2}x^2) \times (1 - nx + \frac{n(n+1)}{2}x^2)
   \]

   First, expand the product term by term:
   \[
   = 1 \times 1 + (-nx) \times 1 + 1 \times (-nx) + (-nx) \times (-nx)
   + \frac{n(n-1)}{2}x^2 \times 1 + 1 \times \frac{n(n+1)}{2}x^2
   \]

   Simplifying:
   \[
   = 1 - 2nx + n^2x^2 + \frac{n(n-1)}{2}x^2 + \frac{n(n+1)}{2}x^2
   \]

5. **Combine the \(x^2\) terms**:
   \[
   x^2 \text{ coefficient} = n^2 + \frac{n(n-1)}{2} + \frac{n(n+1)}{2}
   \]
   Simplify the expression:
   \[
   n^2 + \frac{n(n-1) + n(n+1)}{2} = n^2 + \frac{n(2n)}{2} = n^2 + n^2 = 2n^2
   \]

6. **Final result**:
   \[
   \left(\frac{1 - x}{1 + x}\right)^n = 1 - 2nx + 2n^2x^2
   \]

Thus, the expansion of \(\left(\frac{1 - x}{1 + x}\right)^n\) in ascending powers of \(x\) up to and including the term in \(x^2\) is:
\[
1 - 2nx + 2n^2x^2
\]

Would you like further clarification or more details?

### Related Questions:
1. How does the binomial theorem help in expanding expressions?
2. Can this method be applied to other fractional powers, such as \(\left(\frac{1+x}{1-x}\right)^n\)?
3. What happens to the expansion if \(n\) is a specific integer, like 2 or 3?
4. How do the higher-order terms in \(x^3, x^4\), etc., look for this expansion?
5. Why do we need to expand up to the \(x^2\) term specifically?

**Tip:** In expansions involving fractions, separating the terms and expanding them individually can simplify the multiplication process.

Expand (1 - x)/(1 + x)^n in ascending powers of x up to and including the term in x^2.

Explore how to expand (1 - x)/(1 + x)^n in ascending powers of x up to and including the term in x^2 using the binomial theorem. This detailed solution covers key algebraic concepts and polynomial expansion.

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