(i) Use the binomial expansion formula: \((1 + ax)^n = 1 + n(ax) + \frac{n(n-1)}{2!}(ax)^2 + \ldots\)

For \((1 - 2x^2)^8\), the first three terms are:

\(1 + 8(-2x^2) + \binom{8}{2}(-2x^2)^2\)

\(= 1 - 16x^2 + 112x^4\)

(ii) Expand \((2 - x^2)(1 - 16x^2 + 112x^4)\).

The coefficient of \(x^4\) is found by considering:

\((2 \times 112) - (-16) = 240\)

(a) To expand \((2 + 3x)^4\), use the binomial theorem:

\((2 + 3x)^4 = \sum_{k=0}^{4} \binom{4}{k} (2)^{4-k} (3x)^k\).

The first three terms are:

\(\binom{4}{0} (2)^4 (3x)^0 = 16\),

\(\binom{4}{1} (2)^3 (3x)^1 = 96x\),

\(\binom{4}{2} (2)^2 (3x)^2 = 216x^2\).

Thus, the first three terms are \(16 + 96x + 216x^2\).

(b) To expand \((1 - 2x)^5\), use the binomial theorem:

\((1 - 2x)^5 = \sum_{k=0}^{5} \binom{5}{k} (1)^{5-k} (-2x)^k\).

The first three terms are:

\(\binom{5}{0} (1)^5 (-2x)^0 = 1\),

\(\binom{5}{1} (1)^4 (-2x)^1 = -10x\),

\(\binom{5}{2} (1)^3 (-2x)^2 = 40x^2\).

Thus, the first three terms are \(1 - 10x + 40x^2\).

(c) To find the coefficient of \(x^2\) in \((2 + 3x)^4 (1 - 2x)^5\), consider the products that result in \(x^2\):

\((16)(40) - (10)(96) + (1)(216)\).

Calculate:

\(640 - 960 + 216 = -104\).

Thus, the coefficient of \(x^2\) is \(-104\).

(i) To find the first three terms in the expansion of \(\left( x - \frac{2}{x} \right)^6\), we use the binomial theorem:

\(\left( x - \frac{2}{x} \right)^6 = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} \left( -\frac{2}{x} \right)^k\)

Calculating the first three terms:

For \(k = 0\): \(\binom{6}{0} x^6 = x^6\)

For \(k = 1\): \(\binom{6}{1} x^5 \left( -\frac{2}{x} \right) = -12x^4\)

For \(k = 2\): \(\binom{6}{2} x^4 \left( -\frac{2}{x} \right)^2 = 60x^2\)

Thus, the first three terms are \(x^6 - 12x^4 + 60x^2\).

(ii) To find the coefficient of \(x^4\) in the expansion of \((1 + x^2) \left( x - \frac{2}{x} \right)^6\), we multiply the expansion from part (i) by \((1 + x^2)\):

\((1 + x^2)(x^6 - 12x^4 + 60x^2)\)

We need the terms that result in \(x^4\):

\(1 \times (-12x^4) = -12x^4\)

\(x^2 \times 60x^2 = 60x^4\)

Adding these gives \(-12 + 60 = 48\).

Thus, the coefficient of \(x^4\) is 48.

(i) The binomial expansion of \((1 + ax)^5\) is given by:

\(1 + 5(ax) + \frac{5 \times 4}{2}(ax)^2 + \cdots\)

\(= 1 + 5ax + 10a^2x^2\)

(ii) For the expansion of \((1 - 2x)(1 + ax)^5\), the term in \(x\) is:

\((1)(5ax) + (-2x)(1) = 5ax - 2x\)

Setting the coefficient of \(x\) to zero:

\(5a - 2 = 0\)

\(a = \frac{2}{5}\)

(iii) The coefficient of \(x^2\) in \((1 - 2x)(1 + ax)^5\) is:

\((-2x)(5ax) + (1)(10a^2x^2) = -10ax + 10a^2x^2\)

Substituting \(a = \frac{2}{5}\):

\(-10 \times \frac{2}{5} + 10 \left(\frac{2}{5}\right)^2 = -4 + 1.6 = -2.4\)

(i) To find the first 3 terms in the expansion of \(\left( 2x - \frac{3}{x} \right)^5\), we use the binomial theorem:

\(\left( 2x - \frac{3}{x} \right)^5 = \sum_{k=0}^{5} \binom{5}{k} (2x)^{5-k} \left(-\frac{3}{x}\right)^k\).

The first three terms are:

For \(k=0\): \(\binom{5}{0} (2x)^5 \left(-\frac{3}{x}\right)^0 = 32x^5\).

For \(k=1\): \(\binom{5}{1} (2x)^4 \left(-\frac{3}{x}\right)^1 = -240x^3\).

For \(k=2\): \(\binom{5}{2} (2x)^3 \left(-\frac{3}{x}\right)^2 = 720x\).

Thus, the first three terms are \(32x^5 - 240x^3 + 720x\).

(ii) To find the coefficient of \(x\) in the expansion of \(\left( 1 + \frac{2}{x^2} \right) \left( 2x - \frac{3}{x} \right)^5\), we multiply the first three terms of the expansion by \(1 + \frac{2}{x^2}\):

\((1 + \frac{2}{x^2})(32x^5 - 240x^3 + 720x)\).

The coefficient of \(x\) comes from:

\(1 \times 720x = 720x\) and \(\frac{2}{x^2} \times (-240x^3) = -480x\).

Adding these gives \(720 - 480 = 240\).

Thus, the coefficient of \(x\) is 240.