(a)(i) Expand \((1 + 2x)^5\) using the binomial theorem:

\((1 + 2x)^5 = 1 + 5(2x) + \frac{5 \times 4}{2}(2x)^2 = 1 + 10x + 40x^2\).

(a)(ii) Expand \((1 - ax)^6\) using the binomial theorem:

\((1 - ax)^6 = 1 - 6(ax) + \frac{6 \times 5}{2}(ax)^2 = 1 - 6ax + 15a^2x^2\).

(b) To find the coefficient of \(x^2\) in \((1 + 2x)^5(1 - ax)^6\), consider the terms that contribute to \(x^2\):

1. \(x^2\) term from \((1 + 2x)^5\) and constant term from \((1 - ax)^6\): \(40x^2 \times 1 = 40x^2\).

2. \(x\) term from \((1 + 2x)^5\) and \(x\) term from \((1 - ax)^6\): \(10x \times (-6ax) = -60ax^2\).

3. Constant term from \((1 + 2x)^5\) and \(x^2\) term from \((1 - ax)^6\): \(1 \times 15a^2x^2 = 15a^2x^2\).

Combine these to find the total coefficient of \(x^2\):

\(40 - 60a + 15a^2 = -5\).

Simplify and solve the quadratic equation:

\(15a^2 - 60a + 40 = -5\).

\(15a^2 - 60a + 45 = 0\).

Divide by 15:

\(a^2 - 4a + 3 = 0\).

Factorize:

\((a - 1)(a - 3) = 0\).

Thus, \(a = 1\) and \(a = 3\).

(a) To find the first three terms in the expansion of \((1 + ax)^6\), we use the binomial theorem:

\((1 + ax)^6 = 1 + \binom{6}{1}(ax) + \binom{6}{2}(ax)^2 + \ldots\)

Calculating the first three terms:

\(1 + 6ax + 15a^2x^2\)

(b) To find the coefficient of \(x^2\) in the expansion of \((1 - 3x)(1 + ax)^6\), we first find the relevant terms from each factor:

The expansion of \((1 + ax)^6\) gives the terms \(1, 6ax, 15a^2x^2\).

We need the coefficient of \(x^2\) from the product:

\((1)(15a^2x^2) + (-3x)(6ax) = 15a^2x^2 - 18ax^2\)

Setting the coefficient equal to \(-3\):

\(15a^2 - 18a = -3\)

Rearranging gives:

\(15a^2 - 18a + 3 = 0\)

Factoring the quadratic:

\(3(5a^2 - 6a + 1) = 0\)

\((3)(a-1)(5a-1) = 0\)

Thus, \(a = 1\) or \(a = \frac{1}{5}\).

(a) To find the coefficient of \(x^2\) in the expansion of \((4 + 2x)(2 - ax)^5\), consider the terms that contribute to \(x^2\):

- From \((4)(2 - ax)^5\), the term contributing to \(x^2\) is \(-80ax\).

- From \((2x)(2 - ax)^5\), the term contributing to \(x^2\) is \(80a^2x^2\).

The coefficient of \(x^2\) is given by:

\(320a^2 - 160a = -15\)

Rearranging gives:

\(320a^2 - 160a + 15 = 0\)

Solving this quadratic equation using the quadratic formula:

\(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

where \(a = 320\), \(b = -160\), \(c = 15\).

\(a = \frac{160 \pm \sqrt{(-160)^2 - 4 \times 320 \times 15}}{2 \times 320}\)

\(a = \frac{160 \pm \sqrt{25600 - 19200}}{640}\)

\(a = \frac{160 \pm \sqrt{6400}}{640}\)

\(a = \frac{160 \pm 80}{640}\)

\(a = \frac{240}{640} \text{ or } a = \frac{80}{640}\)

\(a = \frac{3}{8} \text{ or } a = \frac{1}{8}\)

(b) Given that there is only one value of \(a\) for which the coefficient of \(x^2\) is \(k\), we set the discriminant of the quadratic equation to zero:

\(160^2 - 4 \times 320 \times (15 - k) = 0\)

Solving for \(k\):

\(25600 - 1280(15 - k) = 0\)

\(25600 - 19200 + 1280k = 0\)

\(6400 + 1280k = 0\)

\(k = -\frac{6400}{1280}\)

\(k = -20\)

Substituting \(k = -20\) back into the quadratic equation gives:

\(320a^2 - 160a + 20 = 0\)

Solving this gives \(a = \frac{1}{4}\).

First, find the coefficient of \(x\) in the expansion of \(\left(4x + \frac{10}{x}\right)^3\).

The term that gives \(x\) is \(\binom{3}{2} (4x)^2 \left(\frac{10}{x}\right)^1 = 3 \times 16x^2 \times \frac{10}{x} = 480x\).

Thus, \(p = 480\).

Next, find the coefficient of \(\frac{1}{x}\) in the expansion of \(\left(2x + \frac{k}{x^2}\right)^5\).

The term that gives \(\frac{1}{x}\) is \(\binom{5}{3} (2x)^3 \left(\frac{k}{x^2}\right)^2 = 10 \times 8x^3 \times \frac{k^2}{x^4} = 80k^2 \times \frac{1}{x}\).

Thus, \(q = 80k^2\).

Given \(p = 6q\), we have:

\(480 = 6 \times 80k^2\)

\(480 = 480k^2\)

\(k^2 = 1\)

Therefore, \(k = \pm 1\).

First, find the coefficient of \(x^3\) in \((1 - 2x)^5\). The general term in the binomial expansion of \((1 - 2x)^5\) is given by:

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

For \(x^3\), set \(r = 3\):

\(\binom{5}{3} (-2)^3 x^3 = 10(-8)x^3 = -80x^3\).

Thus, the coefficient of \(x^3\) in \((1 - 2x)^5\) is \(-80\).

Next, find the coefficient of \(x^2\) in \((1 - 2x)^5\) to combine with \(kx\) from \((1 + kx)\):

For \(x^2\), set \(r = 2\):

\(\binom{5}{2} (-2)^2 x^2 = 10(4)x^2 = 40x^2\).

Thus, the coefficient of \(x^2\) in \((1 - 2x)^5\) is \(40\).

The coefficient of \(x^3\) in \((1 + kx)(1 - 2x)^5\) is given by:

\(40k - 80 = 20\).

Solving for \(k\):

\(40k - 80 = 20\)

\(40k = 100\)

\(k = \frac{100}{40} = \frac{5}{2}\).