https://file.aiquickdraw.com/mathbot/file/1725422510022v9m3ks8d.jpg

Let's solve the two parts of the problem shown in the image.

### Part (i)
We need to find the first three terms in the expansion of \((1 - 2x)^8\) in ascending powers of \(x\).

To expand \((1 - 2x)^8\), we use the Binomial Theorem:

\[
(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k
\]

Here, \(a = 1\), \(b = -2x\), and \(n = 8\). The first three terms of the expansion will be for \(k = 0\), \(k = 1\), and \(k = 2\).

1. **Term for \(k = 0\)**:
   \[
   \binom{8}{0} \cdot 1^{8-0} \cdot (-2x)^0 = 1 \cdot 1 \cdot 1 = 1
   \]

2. **Term for \(k = 1\)**:
   \[
   \binom{8}{1} \cdot 1^{8-1} \cdot (-2x)^1 = 8 \cdot 1 \cdot (-2x) = -16x
   \]

3. **Term for \(k = 2\)**:
   \[
   \binom{8}{2} \cdot 1^{8-2} \cdot (-2x)^2 = \frac{8 \times 7}{2 \times 1} \cdot 1 \cdot 4x^2 = 28 \cdot 4x^2 = 112x^2
   \]

Thus, the first three terms in the expansion of \((1 - 2x)^8\) are:
\[
1 - 16x + 112x^2
\]

### Part (ii)
Now, we need to find the coefficient of \(x^2\) in the expansion of the expression:

\[
\frac{7 + 3x}{\sqrt{1 - 2x}}
\]

First, express \(\sqrt{1 - 2x}\) as a binomial series. Using the Binomial Theorem for negative exponents:

\[
(1 - 2x)^{-\frac{1}{2}} = \sum_{k=0}^{\infty} \binom{-\frac{1}{2}}{k} (-2x)^k
\]

The first few terms of the expansion are:
\[
(1 - 2x)^{-\frac{1}{2}} = 1 + \left(-\frac{1}{2}\right)(-2x) + \frac{-\frac{1}{2} \cdot -\frac{3}{2}}{2!}(-2x)^2 + \dots
\]
\[
= 1 + x + \frac{3x^2}{2} + \dots
\]

Now, consider the product:
\[
\frac{7 + 3x}{\sqrt{1 - 2x}} = (7 + 3x)(1 + x + \frac{3x^2}{2} + \dots)
\]

Expanding and collecting the terms up to \(x^2\):
\[
= 7(1 + x + \frac{3x^2}{2}) + 3x(1 + x)
= 7 + 7x + \frac{21x^2}{2} + 3x + 3x^2
\]
\[
= 7 + 10x + \left(\frac{21}{2} + 3\right)x^2
= 7 + 10x + \frac{27x^2}{2}
\]

Therefore, the coefficient of \(x^2\) is \(\frac{27}{2}\).

---

Would you like more details or have any questions?

Here are five related questions you might find useful:
1. How do we expand \((1 + x)^n\) for large values of \(n\)?
2. How does the Binomial Theorem change for negative and fractional exponents?
3. What is the general form of the Taylor series expansion for \(\sqrt{1 - x}\)?
4. How would the expansion change if the exponent in \(\sqrt{1 - 2x}\) was \(-\frac{3}{2}\)?
5. Can you explain the convergence criteria for the series expansions used here?

**Tip:** When expanding functions involving radicals, remember that the series may converge only for certain values of \(x\), typically \(|x| < 1\).

Find the first three terms in the expansion of (1 - 2x)^8 in ascending powers of x, where |x| < 1. Hence, find the coefficient of x^2 in the expansion of (7 + 3x)/√(1 - 2x).

This problem demonstrates how to expand (1 - 2x)^8 using the Binomial Theorem and how to calculate the coefficient of x^2 in a complex expression involving radicals. Ideal for students in Grades 11-12.

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