22. (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 . The first two terms in the expansion of \( (2+x)\left(2-\frac{x}{8}\right)^{n} \), in ascending powers of \( x \), are \( a+b x^{2} \), where \( a \) and \( b \) are constants.

We begin by using the Binomial Theorem to expand \[ \left(2-\frac{x}{8}\right)^n, \] where \( n \) is a positive integer with \( n>2 \). The general term is given by \[ \binom{n}{k} \, (2)^{n-k}\left(-\frac{x}{8}\right)^k. \] For the first three terms (with \( k=0,1,2 \)) we have: 1. For \( k=0 \): \[ \binom{n}{0}\,(2)^n\left(-\frac{x}{8}\right)^0 = 2^n. \] 2. For \( k=1 \): \[ \binom{n}{1}\,(2)^{n-1}\left(-\frac{x}{8}\right) = -\frac{n\,2^{n-1}}{8}\,x. \] 3. For \( k=2 \): \[ \binom{n}{2}\,(2)^{n-2}\left(-\frac{x}{8}\right)^2 = \binom{n}{2}\,2^{n-2}\frac{x^2}{64}. \] (Note that the negative sign is squared so it becomes positive.) Thus, the first three terms in ascending powers of \( x \) are \[ 2^n \;-\; \frac{n\,2^{n-1}}{8}\,x \;+\; \binom{n}{2}\,\frac{2^{n-2}}{64}\,x^2. \] Next, we consider the expansion of \[ (2+x)\left(2-\frac{x}{8}\right)^n. \] Multiplying \((2+x)\) by the expansion of \(\left(2-\frac{x}{8}\right)^n\) (keeping only terms up to \(x^2\)) we get: \[ (2+x)\left(2^n - \frac{n\,2^{n-1}}{8}x + \binom{n}{2}\frac{2^{n-2}}{64}x^2\right). \] We now compute the product term‐by‐term. 1. The constant (or \( x^0 \)) term comes only from \[ 2\cdot2^n = 2^{n+1}. \] 2. The \( x^1 \) term receives contributions from: - \(2\) times the coefficient of \( x \) in the expansion: \[ 2\left(-\frac{n\,2^{n-1}}{8}\right) = -\frac{n\,2^n}{8}, \] - and \( x \) times the constant term: \[ 2^n. \] So the total coefficient of \( x \) is: \[ -\frac{n\,2^n}{8} + 2^n = 2^n\left(1 - \frac{n}{8}\right). \] 3. The \( x^2 \) term receives contributions from: - \(2\) times the coefficient of \( x^2 \) in the expansion: \[ 2 \cdot \binom{n}{2}\frac{2^{n-2}}{64} = \binom{n}{2}\frac{2^{n-1}}{64}, \] - and \( x \) times the coefficient of \( x \) in the expansion: \[ x \cdot \left(-\frac{n\,2^{n-1}}{8}x\right) = -\frac{n\,2^{n-1}}{8}\,x^2. \] Thus the total coefficient of \( x^2 \) is: \[ \binom{n}{2}\frac{2^{n-1}}{64} - \frac{n\,2^{n-1}}{8}. \] We are told that the first two terms of the expansion of \((2+x)(2-\frac{x}{8})^n\) are of the form \[ a + bx^2, \] which means that the coefficient of \( x \) must vanish. Hence, we set \[ 2^n\left(1 - \frac{n}{8}\right) = 0. \] Since \(2^n \neq 0\), it follows that \[ 1-\frac{n}{8} = 0 \quad\Longrightarrow\quad n = 8. \] Now, with \( n = 8 \) we find the constants \( a \) and \( b \). 1. The constant term is: \[ a = 2^{8+1} = 2^9 = 512. \] 2. To find \( b \) (the coefficient of \( x^2 \)): First, compute the coefficient of \( x^2 \) in the expansion of \(\left(2-\frac{x}{8}\right)^8\). For \( n=8 \), \[ \binom{8}{2} = \frac{8\cdot7}{2} = 28, \] and since the \( x^2 \) term is \[ 28\,\frac{2^{8-2}}{64}\,x^2 = 28\,\frac{2^6}{64}\,x^2 = 28\,\frac{64}{64}\,x^2 = 28x^2, \] we now compute the contributions to the \( x^2 \) term: - From \(2\) multiplied by the \(x^2\) term: \[ 2\cdot 28x^2 = 56x^2, \] - From \(x\) multiplied by the \(x\) term of the expansion of \(\left(2-\frac{x}{8}\right)^8\). The \( x \) term was: \[ -\frac{8\,2^{7}}{8}\,x = -2^{7}\,x = -128x, \] so multiplying by \( x \) gives \[ -128x^2. \] Adding these contributions gives: \[ b = 56 - 128 = -72. \] Thus, the answers are: For (i) the first three terms in the expansion of \[ \left(2-\frac{x}{8}\right)^n \] are \[ 2^n \;-\; \frac{n\,2^{n-1}}{8}\,x \;+\; \binom{n}{2}\frac{2^{n-2}}{64}\,x^2, \] and the first two terms of \[ (2+x)\left(2-\frac{x}{8}\right)^n \] (with cancellation of the \( x \) term) occur when \( n = 8 \) and are given by \[ 512 - 72x^2. \]