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Let's break down and solve the given problem step by step.

### Part (i): Expand \( \left( \frac{1 - x}{1 + x} \right)^n \) in ascending powers of \(x\) up to and including the term in \(x^2\).

#### Step 1: Express the function as a product
We can rewrite \( \frac{1 - x}{1 + x} \) as:

\[
\frac{1 - x}{1 + x} = (1 - x)(1 + x)^{-1}
\]

Now, expand \( (1 + x)^{-1} \) using the binomial series for negative exponents:
\[
(1 + x)^{-1} = 1 - x + x^2 - x^3 + \cdots
\]

#### Step 2: Multiply and simplify
Now, multiply this expansion by \(1 - x\):

\[
(1 - x)(1 - x + x^2 - x^3 + \cdots) = 1 - x - x + x^2 + x^2 - x^3 + \cdots
\]

Simplifying terms up to \(x^2\), we get:

\[
1 - 2x + 2x^2
\]

#### Step 3: Raise to the power \(n\)
Now, raise the result \( (1 - 2x + 2x^2) \) to the power of \(n\). Using the binomial theorem:

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

This simplifies to:

\[
1 - 2nx + n(n-1)x^2
\]

Thus, the expansion up to the term in \(x^2\) is:

\[
1 - 2nx + n(n-1)x^2
\]

---

### Part (ii): State the set of values of \(x\) for which the series expansion is valid.

The expansion of \( (1 + x)^{-1} \) is valid for \( |x| < 1 \). Therefore, the series expansion of \( \left( \frac{1 - x}{1 + x} \right)^n \) is valid for:

\[
|x| < 1
\]

---

### Part (iii): Find an approximation to the fourth root of \( \frac{19}{21} \), in the form \( \frac{p}{q} \), where \(p\) and \(q\) are positive integers with no common factors.

We can use the expansion we derived in part (i) to approximate \( \left( \frac{19}{21} \right)^{1/4} \).

#### Step 1: Express \( \frac{19}{21} \) as \( 1 - x \)
Let:
\[
\frac{19}{21} = 1 - x
\]

So:
\[
x = \frac{2}{21}
\]

#### Step 2: Use the binomial approximation
Now, use the approximation for \( (1 - x)^{1/4} \):

\[
(1 - x)^{1/4} \approx 1 - \frac{1}{4}x
\]

Substituting \( x = \frac{2}{21} \):

\[
(1 - \frac{2}{21})^{1/4} \approx 1 - \frac{1}{4} \times \frac{2}{21} = 1 - \frac{2}{84} = 1 - \frac{1}{42}
\]

Thus:

\[
(1 - \frac{2}{21})^{1/4} \approx \frac{41}{42}
\]

#### Step 3: Ensure \( p \) and \( q \) are in their simplest form
The approximation is \( \frac{41}{42} \), and since 41 and 42 have no common factors, \( p = 41 \) and \( q = 42 \).

---

### Summary of Answers:
1. The expansion of \( \left( \frac{1 - x}{1 + x} \right)^n \) up to \(x^2\) is:
   \[
   1 - 2nx + n(n-1)x^2
   \]
   
2. The series expansion is valid for \( |x| < 1 \).

3. The approximation to the fourth root of \( \frac{19}{21} \) is \( \frac{41}{42} \).

---

Would you like further details or clarifications on any part of this solution? Here are some related questions to expand upon the topic:

1. How is the binomial expansion derived for negative exponents?
2. What other methods can be used to approximate roots of fractions?
3. How does the range of validity change when expanding other functions?
4. What is the full binomial expansion of \( (1 + x)^{-n} \)?
5. How can you apply the expansion method to solve real-world problems?

**Tip**: When approximating roots or fractions, always check if the series expansion used converges in the desired interval of \(x\).

Expand (1 - x)/(1 + x) to the power of n in ascending powers of x up to and including the term in x^2. State the set of values of x for which the series expansion is valid. Find an approximation to the fourth root of 19/21 in the form p/q, where p and q are positive integers with no common factors.

Learn how to expand (1 - x)/(1 + x) to the power of n using binomial expansion up to the term in x^2. This problem involves algebraic concepts such as series, sequences, and binomial theorems. Suitable for Grades 11-12.

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