Let's go through the question step-by-step.

Part (a)

We are asked to find the first four terms of the binomial expansion of:

$(1 + 8x)^{\frac{1}{2}}$

Formula for Binomial Expansion:

For small $x$, the binomial expansion of $(1 + ax)^n$ is:

$(1 + ax)^n = 1 + n \cdot ax + \frac{n(n - 1)}{2!} \cdot (ax)^2 + \frac{n(n - 1)(n - 2)}{3!} \cdot (ax)^3 + \dots$

In our case:

• $n = \frac{1}{2}$
• $a = 8$

Now let's find the first four terms in ascending powers of $x$.

First Term:

This is always $1$.

$\text{First term} = 1$

Second Term:

$\frac{1}{2} \cdot (8x) = 4x$

Third Term:

$\frac{\frac{1}{2} \cdot \left(\frac{1}{2} - 1\right)}{2!} \cdot (8x)^2$

Evaluating step-by-step:

• $\frac{1}{2} - 1 = -\frac{1}{2}$
• $\frac{\frac{1}{2} \cdot -\frac{1}{2}}{2} = -\frac{1}{8}$
• $(8x)^2 = 64x^2$

Thus: $\text{Third term} = -\frac{1}{8} \cdot 64x^2 = -8x^2$

Fourth Term:

$\frac{\frac{1}{2} \cdot \left(\frac{1}{2} - 1\right) \cdot \left(\frac{1}{2} - 2\right)}{3!} \cdot (8x)^3$

Evaluating step-by-step:

• $\frac{1}{2} - 2 = -\frac{3}{2}$
• $\frac{\frac{1}{2} \cdot -\frac{1}{2} \cdot -\frac{3}{2}}{6} = \frac{3}{16}$
• $(8x)^3 = 512x^3$

Thus: $\text{Fourth term} = \frac{3}{16} \cdot 512x^3 = 96x^3$

First Four Terms:

The first four terms in ascending powers of $x$ are:

$1 + 4x - 8x^2 + 96x^3$

Part (b)

We need to explain how $x = \frac{1}{32}$ could be used to approximate $\sqrt{5}$ using the expansion.

Key Idea:

The given expression is $(1 + 8x)^{\frac{1}{2}}$. If we let $x = \frac{1}{32}$, then:

$1 + 8x = 1 + 8 \cdot \frac{1}{32} = 1 + \frac{1}{4} = 1.25$

Thus, the expansion is helping us approximate:

$(1.25)^{\frac{1}{2}}$

Since $1.25 = \frac{5}{4}$, this is equivalent to:

$\sqrt{5} \div 2$

So, by evaluating the expansion for $x = \frac{1}{32}$, we can find an approximation for $\sqrt{5} / 2$. Multiplying this result by 2 will give us an approximation for $\sqrt{5}$.

Do you need further explanation or have any questions?

Here are 5 related questions you could explore next:

1. How does the accuracy of the binomial expansion improve with more terms?
2. Why is the binomial expansion useful for approximating square roots?
3. What conditions need to be satisfied for a binomial expansion to converge?
4. How does the error behave in a truncated binomial expansion?
5. Can this method be extended to cube roots or higher roots?

Tip: Binomial expansion is most accurate when $x$ is small (close to 0). This is because higher-order terms become smaller and have less influence.