3. (a) 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 4 - 2012 - Paper 4

To find the first four terms of the binomial expansion, we use the Binomial Theorem:

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

In this case, let:

• $a = 1$
• $b = \frac{x}{4}$
• $n = 8$

Thus, the first four terms are:

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

Therefore, the first four terms of the expansion are: $1 + 2x + \frac{7}{4}x^2 + \frac{7}{8}x^3$