Base Case: For \(n = 1\), \(6^4 + 38^1 - 2 = 1332\), which is divisible by 74.

Inductive Step: Assume \(6^{4k} + 38^k - 2\) is divisible by 74 for some positive integer \(k\).

Then, consider \(6^{4(k+1)} + 38^{k+1} - 2\):

\(= (1295 + 1) \times 6^{4k} + (37 + 1) \times 38^k - 2\)

This expression is divisible by 74 because \(1295 \times 6^{4k} + 37 \times 38^k\) is divisible by 74.

Hence, by induction, \(6^{4n} + 38^n - 2\) is divisible by 74 for every positive integer \(n\).

(a) Expand \(r(r+1)(3r+4)\) to get \(3r^3 + 7r^2 + 4r\).

Use standard summation formulae:

\(\sum_{r=1}^{N} r^3 = \left( \frac{N(N+1)}{2} \right)^2, \quad \sum_{r=1}^{N} r^2 = \frac{N(N+1)(2N+1)}{6}, \quad \sum_{r=1}^{N} r = \frac{N(N+1)}{2}.\)

Substitute these into the expanded expression:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Simplify to get:

\(\frac{1}{12}N(9N^3 + 46N^2 + 75N + 38) = \frac{1}{12}N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Substitute into the series:

\(\sum_{r=1}^{N} \left( \frac{4}{r} - \frac{1}{r+1} \right) \left( \frac{1}{4} \right)^{r+1}.\)

Write out terms and simplify using the method of differences:

\(\left( \frac{1}{4^2} \right) \left( 4 - \frac{1}{2} \right) + \left( \frac{1}{4^3} \right) \left( 4 - \frac{1}{3} \right) + \ldots + \left( \frac{1}{4^{N+1}} \right) \left( 4 - \frac{1}{N+1} \right).\)

The series telescopes to:

\(\frac{1}{4} \left( 1 - \frac{1}{N+1} \right).\)

(c) As \(N \to \infty\), the series becomes:

\(\frac{1}{4}.\)

(a) The matrix \(\mathbf{M}\) represents a stretch followed by a rotation. The stretch is parallel to the x-axis with a scale factor of 14. The rotation is \(\frac{\pi}{3}\) anticlockwise about the origin.

(b) \(\mathbf{M}^{-1} = \begin{pmatrix} \frac{1}{14} & 0 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{2} & \frac{1}{2}\sqrt{3} \\ -\frac{1}{2}\sqrt{3} & \frac{1}{2} \end{pmatrix}\)

(c) The equations of the invariant lines are \(y = 2\sqrt{3}x\) and \(y = \frac{\sqrt{3}}{3}x\).

(d) Given \(28 = 14 \times |\triangle ABC|\), the area of triangle \(ABC\) is \(2 \text{ cm}^2\).

(a) To find the equation of the plane ABC, we first find the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (2\mathbf{i} + 4\mathbf{j} - \mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = 2\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) - (2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}) = -5\mathbf{i} - 5\mathbf{j}\)

The normal vector to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 2 & -5 \\ -5 & -5 & 0 \end{vmatrix} = -5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k}\)

Using point \(A(2, 2, 4)\), the equation of the plane is:

\(-5(x - 2) + 5(y - 2) + 2(z - 4) = 0\)

Simplifying gives \(-5x + 5y + 2z = 8\).

(b) The perpendicular distance from point D to the plane is given by:

\(\frac{|8 - (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) \cdot (-5\mathbf{i} + 5\mathbf{j} + 2\mathbf{k})|}{\sqrt{(-5)^2 + 5^2 + 2^2}} = \frac{7}{\sqrt{54}} = 0.953\)

(c) The direction vector \(\overrightarrow{CD}\) is:

\(\overrightarrow{CD} = (2\mathbf{i} + \mathbf{j} + 3\mathbf{k}) - (-3\mathbf{i} - 3\mathbf{j} + 4\mathbf{k}) = 5\mathbf{i} + 4\mathbf{j} - \mathbf{k}\)

Using the formula for the shortest distance between two skew lines:

\(\frac{|\overrightarrow{AB} \cdot (\overrightarrow{CD} \times \overrightarrow{AC})|}{|\overrightarrow{CD} \times \overrightarrow{AC}|} = \frac{35}{\sqrt{1049}} = 1.08\)

(a) The vertical asymptote is at \(x = -2\). The oblique asymptote is found by dividing \(x^2 + ax + 1\) by \(x + 2\), giving \(y = x + a - 2\).

(b) Differentiate \(y = \frac{x^2 + ax + 1}{x + 2}\) using the quotient rule: \(\frac{dy}{dx} = \frac{(x+2)(2x+a) - (x^2+ax+1)}{(x+2)^2}\). Simplifying gives \(\frac{dy}{dx} = \frac{x^2 + 4x + 2a - 1}{(x+2)^2}\). Setting the numerator to zero gives \(x^2 + 4x + 2a - 1 = 0\). The discriminant is \(16 - 4(2a-1) = 20 - 8a\), which is less than zero for \(a > \frac{5}{2}\), so no stationary points exist.

(c) The curve intersects the y-axis at \((0, \frac{1}{2})\). The asymptotes are \(x = -2\) and \(y = x + a - 2\).

(d)(i) Sketch the curve \(y = \left| \frac{x^2 + ax + 1}{x + 2} \right|\) with the same asymptotes and a reflection in the x-axis where the original curve is negative.

(d)(ii) Draw the line \(y = a\) on the sketch.

(d)(iii) Solve \(\frac{x^2 + ax + 1}{x + 2} = a\) and \(\frac{x^2 + ax + 1}{x + 2} = -a\). This gives the quadratic \(x^2 + 1 - 2a = 0\) or \(x^2 + 2ax + 1 = 0\). The roots are \(-a - \sqrt{a^2 - 2a - 1} < x < -a + \sqrt{a^2 - 2a - 1}\). Solving for \(a\) gives \(a = 5\).