(a) We start with \(\sum_{r=1}^{n} r(r+1)(r+2) = \sum_{r=1}^{n} (r^3 + 3r^2 + 2r)\).

Using standard summation formulae:

\(\sum_{r=1}^{n} r^3 = \left( \frac{n(n+1)}{2} \right)^2\)

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

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

Substituting these into the expression:

\(\frac{1}{4}n(n+1)^2 + \frac{1}{2}n(n+1)(2n+1) + n(n+1)\)

Factorising gives \(\frac{1}{4}n(n+1)(n+2)(n+3)\).

(b) Express \(\frac{1}{r(r+1)(r+2)}\) in partial fractions:

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

Using the method of differences:

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

This telescopes to:

\(\frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3) + \frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4) + \ldots + \frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)\)

Resulting in \(\frac{1}{4} \left( \frac{1}{n+1} + \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the sum becomes \(\frac{1}{4}\).

(a) Base case: For \(n = 2\), \(a_2 = 2^3\) leading to \(\ln a_2 = 3 \ln 2\), so true when \(n = 2\).

Inductive step: Assume \(\ln a_k \geq 3^{k-1} \ln 2\).

Then \(\ln a_{k+1} = 3 \ln \left( a_k + \frac{1}{a_k} \right) > 3 \ln a_k \geq 3^k \ln 2\).

Thus, true for \(n = k+1\). By induction, true for all integers \(n \geq 2\).

(b) \(3 \ln \left( a_n + \frac{1}{a_n} \right) - \ln a_n > 2 \ln a_n\)

\(> 2 \times 3^{n-1} \ln 2 = 3^{n-1} \ln 4\)

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a one-way stretch with scale factor 3 parallel to the x-axis, followed by a rotation anticlockwise about the origin through an angle \(\theta\).

(b) The matrix \(\mathbf{M} = \begin{pmatrix} 3\cos \theta & -\sin \theta \\ 3\sin \theta & \cos \theta \end{pmatrix}\) transforms \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} 3x\cos \theta - y\sin \theta \\ 3x\sin \theta + y\cos \theta \end{pmatrix}\).

For an invariant line through the origin, \(y = mx\) transforms to \(Y = mX\), leading to the equation:

\(3x\sin \theta + mx\cos \theta = 3mx\cos \theta - m^2x\sin \theta\)

\(m^2 \sin \theta - 2m \cos \theta + 3 \sin \theta = 0\)

Setting the discriminant equal to zero:

\(4\cos^2 \theta - 12\sin^2 \theta = 0\)

Using \(\tan \theta = \pm \frac{1}{\sqrt{3}}\) or \(4\cos^2 \theta - 12(1 - \cos^2 \theta) = 0\) leads to \(16\cos^2 \theta - 12 = 0\) leading to \(\cos \theta = \pm \frac{\sqrt{3}}{2}\).

Thus, \(\theta = \frac{1}{6} \pi\) and \(\theta = \frac{5}{6} \pi\).

(a) The normal vector to the plane \(\Pi\) is found using the cross product of the direction vectors \(\mathbf{i} + \mathbf{k}\) and \(2\mathbf{i} + 3\mathbf{j}\). The determinant is:

This gives the normal vector \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\).

Substitute the point \((-2, 3, 3)\) into the plane equation:

\(-3(-2) + 2(3) + 3(3) = 21\)

Thus, the Cartesian equation is \(-3x + 2y + 3z = 21\).

(b) The line \(l\) has the equation \(\begin{pmatrix} 2 \\ -3 \\ 5 + t \end{pmatrix}\). Substitute into the plane equation:

\(-3(2) + 2(-3) + 3(5 + t) = 21\)

This simplifies to \(t = 6\), giving the point \(\begin{pmatrix} 2 \\ -3 \\ 11 \end{pmatrix}\).

(c) The direction vector of \(l\) is \(\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}\). The normal vector to \(\Pi\) is \(\begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix}\). The dot product is:

\(0 \cdot (-3) + 0 \cdot 2 + 1 \cdot 3 = 3\)

The magnitudes are \(\sqrt{1}\) and \(\sqrt{22}\), so \(\cos \alpha = \frac{3}{\sqrt{22}}\).

The acute angle is \(90^\circ - \alpha = 39.8^\circ\).

(d) The direction vector from \(P\) to the plane is \(\begin{pmatrix} 0 \\ -3 \\ 7 \end{pmatrix}\). The dot product with the normal vector is:

\(\frac{1}{\sqrt{22}} \begin{pmatrix} -2 \\ 3 \\ 5 \end{pmatrix} \cdot \begin{pmatrix} -3 \\ 2 \\ 3 \end{pmatrix} = \frac{18}{\sqrt{22}} = 3.84\).

(a) To find the point furthest from the pole, differentiate \(r\) with respect to \(\theta\):

\(\frac{dr}{d\theta} = 2(\cos \theta) - (1 + \sin \theta) \sin \theta = 0\)

Simplifying gives:

\(1 - s - 2s^2 = 0\)

Solving the quadratic in \(\sin \theta\), we find \(\sin \theta = -1\) or \(\sin \theta = \frac{1}{2}\), leading to \(\theta = \frac{1}{6} \pi\).

The polar coordinates are \(\left( \frac{3}{2} \sqrt{3}, \frac{1}{6} \pi \right)\).

(b) The sketch of C shows a correct shape, tangential to \(\theta = \frac{\pi}{2}\) at the pole, with \(r\) strictly increasing to \(\frac{\pi}{6}\) then decreasing. The curve is in the first quadrant.

(c) The area of the region bounded by C and the initial line is found using:

\(\frac{1}{2} \int_{0}^{\frac{1}{2}\pi} r^2 \, d\theta = \frac{1}{2} \int_{0}^{\frac{1}{2}\pi} 4c^2(1+s)^2 \, d\theta\)

Using double angle formulae and integrating, we obtain:

\(\left[ \frac{5}{4} \theta + \frac{1}{2} \sin 2\theta - \frac{4}{3} \cos^3 \theta - \frac{1}{16} \sin 4\theta \right]_{0}^{\frac{1}{2}\pi}\)

The exact area is \(\frac{5}{8} \pi + \frac{4}{3}\).