(a) Find the binomial expansion of $(4 + 5x)^{\frac{1}{2}}$, $|x| < \frac{4}{5}$ Give each coefficient in its simplest form - Edexcel - A-Level Maths Pure - Question 3 - 2015 - Paper 4

Substituting $x = \frac{1}{10}$ into the expansion: $2 + 5 \left(\frac{1}{10}\right) - \frac{25}{8} \left(\frac{1}{10}\right)^2$ Calculating term by term:

1. The first term is $2$.
2. The second term is $5 \cdot \frac{1}{10} = \frac{1}{2}$.
3. The third term is $-\frac{25}{8} \cdot \frac{1}{100} = -\frac{25}{800} = -\frac{1}{32}$.

Now, combine the terms: $2 + \frac{1}{2} - \frac{1}{32}$ To add these, we convert to a common denominator (32): $= \frac{64}{32} + \frac{16}{32} - \frac{1}{32} = \frac{79}{32}$ Thus, we can write: $\frac{79}{32} = k \sqrt{2}, \text{ where } k = \frac{79}{32 \sqrt{2}}$

Using the expansion from part (a), we substitute $x = \frac{1}{10}$ again: $2 + 5 \cdot \frac{1}{10} - \frac{25}{8} \cdot \left(\frac{1}{10}\right)^2$ This gives us: $2 + \frac{1}{2} - \frac{25}{800}$ Following the same calculations:

1. The first term: $2 = \frac{64}{32}$
2. The second term: $\frac{1}{2} = \frac{16}{32}$
3. The third term: $-\frac{1}{32}$

Combine them: $\frac{64 + 16 - 1}{32} = \frac{79}{32}$ This gives us an approximation for $\sqrt{2}$: $\sqrt{2} \approx \frac{p}{q} = \frac{79}{32}$