(i) Write down the first three terms in the expansion, in ascending powers of \( x \), of \( \left(2-\frac{x}{8}\right)^{n} \), where \( n \) is a positive integer greater than 2 .

To find the first three terms in the expansion of \( \left(2 - \frac{x}{8}\right)^n \), we can use the Binomial Theorem, which states that: \[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] Here, let \( a = 2 \) and \( b = -\frac{x}{8} \). The first three terms (for \( k = 0, 1, 2 \)) will be: 1. For \( k = 0 \): \[ \binom{n}{0} (2)^{n} (-\frac{x}{8})^0 = 2^n \] 2. For \( k = 1 \): \[ \binom{n}{1} (2)^{n-1} (-\frac{x}{8})^{1} = n \cdot 2^{n-1} \cdot \left(-\frac{x}{8}\right) = -\frac{n \cdot 2^{n-1} x}{8} \] 3. For \( k = 2 \): \[ \binom{n}{2} (2)^{n-2} (-\frac{x}{8})^{2} = \frac{n(n-1)}{2} \cdot 2^{n-2} \cdot \left(-\frac{x}{8}\right)^{2} = \frac{n(n-1) \cdot 2^{n-2} x^2}{64} \] Combining these, the first three terms in ascending powers of \( x \) are: \[ 2^n - \frac{n \cdot 2^{n-1} x}{8} + \frac{n(n-1) \cdot 2^{n-2} x^2}{64} \]