Find the first 4 terms of the binomial expansion, in ascending powers of $x$, of $$\left(1+\frac{x}{4}\right)^{8}$$ giving each term in its simplest form - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 4

To find the first four terms of the binomial expansion of (\left(1+\frac{x}{4}\right)^{8}), we can use the binomial theorem, which states that:

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

In this case, let (a = 1), (b = \frac{x}{4}), and (n = 8). The first few terms of the expansion are:

1. For k=0: ( \binom{8}{0} \cdot 1^{8} \cdot \left(\frac{x}{4}\right)^{0} = 1 )
2. For k=1: ( \binom{8}{1} \cdot 1^{7} \cdot \left(\frac{x}{4}\right)^{1} = 8 \cdot \frac{x}{4} = 2x )
3. For k=2: ( \binom{8}{2} \cdot 1^{6} \cdot \left(\frac{x}{4}\right)^{2} = 28 \cdot \frac{x^{2}}{16} = \frac{7}{4}x^{2} )
4. For k=3: ( \binom{8}{3} \cdot 1^{5} \cdot \left(\frac{x}{4}\right)^{3} = 56 \cdot \frac{x^{3}}{64} = \frac{7}{8}x^{3} )

Combining these, the first four terms in ascending order are:

$1 + 2x + \frac{7}{4}x^{2} + \frac{7}{8}x^{3}$