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:

   + \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: $1 - 2nx + 2n^2x^2$

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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.