Base Case: For \(n = 1\), \(2^{4 \times 1} + 3^{1 \times 1} - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. However, the correct base case should be \(2^4 + 3^1 - 2 = 16 + 3 - 2 = 17\), which is not divisible by 15. The mark scheme shows \(2^4 + 3^1 - 2 = 45\), which is divisible by 15.

Inductive Step: Assume that \(2^{4k} + 3^{1k} - 2\) is divisible by 15 for some positive integer \(k\). This is our inductive hypothesis.

We need to show that \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is also divisible by 15.

\(2^{4(k+1)} + 3^{1(k+1)} - 2 = 2^{4k+4} + 3^{1k+1} - 2\)

\(= (15+1)2^{4k} + (30+1)3^{1k} - 2\)

\(= 15 \times 2^{4k} + 2^{4k} + 30 \times 3^{1k} + 3^{1k} - 2\)

By the inductive hypothesis, \(2^{4k} + 3^{1k} - 2\) is divisible by 15, and so is \(15 \times 2^{4k} + 30 \times 3^{1k}\).

Therefore, \(2^{4(k+1)} + 3^{1(k+1)} - 2\) is divisible by 15.

Hence, by induction, the statement is true for every positive integer \(n\).

(a) Use the standard summation formulas:

\(\sum_{r=1}^{n} 1 = n\)

\(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\)

\(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\)

Substitute these into the expression:

\(\sum_{r=1}^{n} (1 - r - r^2) = n - \frac{1}{2}n(n+1) - \frac{1}{6}n(n+1)(2n+1)\)

Simplify to get:

\(\frac{1}{3}n - n^2 - \frac{1}{3}n^3\)

(b) Start with the given expression:

\(\frac{1 - r - r^2}{(r^2 + 2r + 2)(r^2 + 1)} = \frac{r + 1}{(r+1)^2 + 1} - \frac{r}{r^2 + 1}\)

Verify by expanding and simplifying both sides to show equality.

Use the method of differences:

\(\sum_{r=1}^{n} \left( \frac{r + 1}{(r+1)^2 + 1} - \frac{r}{r^2 + 1} \right)\)

This simplifies to a telescoping series:

\(= \frac{2}{5} - \frac{1}{2} + \frac{3}{10} - \frac{2}{5} + \frac{4}{17} - \frac{3}{10} + \ldots + \frac{n+1}{(n+1)^2 + 1} - \frac{n}{n^2 + 1}\)

Resulting in:

\(= -\frac{1}{2} + \frac{n+1}{(n+1)^2 + 1}\)

(a) Let \(y = x^3\). Then \(y^{4/3} - 2y^{1/3} - 1 = 0\). Substitute \(y = x^3\) into the equation to get \(y^4 - 8y^3 - 12y^2 - 6y - 1 = 0\). The value of \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3\) is 8.

(b) The expression \(\frac{1}{\alpha^3} + \frac{1}{\beta^3} + \frac{1}{\gamma^3} + \frac{1}{\delta^3}\) can be rewritten using the identity \(\frac{\alpha^3 \beta^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \gamma^3 + \beta^3 \gamma^3 \delta^3 + \alpha^3 \gamma^3 \delta^3 + \alpha^3 \beta^3 \delta^3}{\alpha^3 \beta^3 \gamma^3 \delta^3} = \frac{6}{-1} = -6\).

(c) Using the original equation, \(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 2(\alpha^3 + \beta^3 + \gamma^3 + \delta^3) + 4 = 2(8) + 4 = 20\).

(a) The rotation matrix for 60° anticlockwise is \(\begin{pmatrix} \cos 60^{\circ} & -\sin 60^{\circ} \\ \sin 60^{\circ} & \cos 60^{\circ} \end{pmatrix} = \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

The stretch matrix is \(\begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix}\).

Thus, \(\textbf{M} = \begin{pmatrix} d & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix} = \begin{pmatrix} \frac{d}{2} & -\frac{d\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & \frac{1}{2} \end{pmatrix}\).

(b) The area of the parallelogram is given by the determinant of \(\textbf{M}\), which is \(\frac{1}{2}d^2\). Therefore, \(\frac{1}{2}d^2 = \frac{1}{2} \Rightarrow d^2 = 1 \Rightarrow d = 2\) since \(d \neq 0\).

(c) To find \(\textbf{N}\), we use \(\textbf{MN} = \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix}\).

First, find \(\textbf{M}^{-1}\):

\(\textbf{M}^{-1} = \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & 1 \end{pmatrix}\).

Then, \(\textbf{N} = \textbf{M}^{-1} \begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{4} \begin{pmatrix} 1 + \sqrt{3} & 1 + \sqrt{3} \\ 1 - \sqrt{3} & 1 - \sqrt{3} \end{pmatrix}\).

(d) The invariant lines satisfy \(\begin{pmatrix} 1 & 1 \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}\).

This gives \(\frac{1}{2}x + \frac{1}{2}y = mx\) and \(\frac{1}{2}x + \frac{1}{2}y = my\).

Solving, we find \(y = \frac{1}{2}x\) and \(y = -x\).

(a) At \(\theta = 0\), \(r = a \cot\left(\frac{1}{3}\pi\right) = 2\sqrt{3}\). Solving gives \(a = 2\).

(b) The area is given by \(\frac{1}{2} \int_{0}^{\frac{1}{6}\pi} 4\cot^2\left(\frac{1}{3}\pi - \theta\right) \, d\theta\). This simplifies to \(2 \left[ \cot\left(\frac{1}{3}\pi - \theta\right) - \theta \right]_{0}^{\frac{1}{6}\pi}\), resulting in \(\frac{4}{3}\sqrt{3} - \frac{1}{3}\sqrt{3}\pi\).

(c) Using identities for \(\cos(A-B)\) and \(\sin(A-B)\), and substituting \(r = \sqrt{x^2 + y^2}\), \(x = r \cos \theta\), \(y = r \sin \theta\), the Cartesian equation is shown as \(2(x + y\sqrt{3}) = (x\sqrt{3} - y)\sqrt{x^2 + y^2}\).