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 3 - 2020 - Paper 1

To find the first four terms of the binomial expansion of

(1 + 8y)^{rac{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:

• $a = 1$
• $b = 8y$
• $n = \frac{1}{2}$

Applying the binomial expansion gives:

1. First term (k=0): inom{\frac{1}{2}}{0} (1)^{\frac{1}{2}} (8y)^0 = 1

2. Second term (k=1): $\binom{\frac{1}{2}}{1} (1)^{\frac{1}{2}} (8y)^1 = \frac{1}{2} \cdot 8y = 4y$

3. Third term (k=2): $\binom{\frac{1}{2}}{2} (1)^{\frac{1}{2}} (8y)^2 = \frac{\frac{1}{2}\left(\frac{1}{2}-1\right)}{2!} (8y)^2 = -\frac{1}{8} \cdot 64y^2 = -8y^2$

4. Fourth term (k=3): $\binom{\frac{1}{2}}{3} (1)^{\frac{1}{2}} (8y)^3 = \frac{\frac{1}{2} \left(\frac{1}{2}-1\right) \left(\frac{1}{2}-2\right)}{3!} (8y)^3 = \frac{-1/2 \cdot -1/2 \cdot -3/2}{6} \cdot 512y^3 = \frac{-64}{6} y^3 = -\frac{32}{3} y^3$

Thus, the first four terms in descending powers of $y$ are:

$1 + 4y - 8y^2 - \frac{32}{3} y^3$