1. (a) Find the first four terms, in ascending powers of $x$, of the binomial expansion of (1 + 8y)^{rac{1}{2}} giving each term in simplest form - Edexcel - A-Level Maths Pure - Question 2 - 2020 - Paper 1

To find the first four terms of the binomial expansion of $(1 + 8y)^{\frac{1}{2}}$, we can use the binomial theorem, which states:

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

In our case, let $a = 1$, $b = 8y$, and $n = \frac{1}{2}$. The coefficients can be computed as follows:

1. First Term (k=0):

   $T_0 = \binom{\frac{1}{2}}{0} (1)^{\frac{1}{2}} (8y)^0 = 1$

2. Second Term (k=1):

   $T_1 = \binom{\frac{1}{2}}{1} (1)^{\frac{1}{2}} (8y)^1 = \frac{1}{2} \cdot 8y = 4y$

3. Third Term (k=2):

   $T_2 = \binom{\frac{1}{2}}{2} (1)^{\frac{1}{2}} (8y)^2 = \frac{\frac{1}{2} \cdot (\frac{1}{2}-1)}{2!} \cdot 64y^2 = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot 32y^2 = -8y^2$

4. Fourth Term (k=3):

   $T_3 = \binom{\frac{1}{2}}{3} (1)^{\frac{1}{2}} (8y)^3 = \frac{\frac{1}{2} \cdot (\frac{1}{2}-1)(\frac{1}{2}-2)}{3!} \cdot 512y^3 = \frac{1}{2} \cdot \left(-\frac{1}{2}\right) \cdot \left(-\frac{3}{2}\right) \cdot \frac{512}{6}y^3 = \frac{128}{3}y^3$

Thus, the first four terms of the expansion are:

$1 + 4y - 8y^2 + \frac{128}{3}y^3$