(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

Simplify to get \(3n^3 - 6n^2 + n\).

(b) Express \(\frac{1}{(2-3r)(5-3r)}\) as partial fractions: \(\frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Write the sum: \(\sum_{r=1}^{n} \frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Observe cancellation: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum converges to \(-\frac{1}{6}\).

(a) Let \(y = x^3 - 1\). Then \(x^3 = y + 1\).

Substitute into the original equation: \((y+1) + 2(y+1)^{1/3} + 1 = 0\).

This simplifies to \(2(y+1)^{1/3} = -y - 2\).

Cube both sides: \(8(y+1) = -(y+2)^3\).

Expand and simplify: \(8y + 8 = -y^3 - 6y^2 - 12y - 8\).

Rearrange to get the cubic equation: \(y^3 + 6y^2 + 20y + 16 = 0\).

(b) Use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Given \(\alpha + \beta + \gamma = 0\), \(\alpha\beta + \beta\gamma + \gamma\alpha = 2\), and \(\alpha\beta\gamma = -1\), calculate \((\alpha^3 - 1)^2 + (\beta^3 - 1)^2 + (\gamma^3 - 1)^2 = 6^2 - 2(20) = -4\).

(c) Use the result from (a) or a valid formula for the sum of cubes: \((\alpha^3 - 1)^3 + (\beta^3 - 1)^3 + (\gamma^3 - 1)^3 = -6(-6) - 20(-6) - 16(3) = 96\).

(a) Base case: For \(n = 1\), \(u_1 = 5 = 6^1 - 1\). Assume it is true for \(n = k\), so \(u_k = 6^k - 1\).

Then \(u_{k+1} = 6(6^k - 1) + 5 = 6^{k+1} - 1\).

So, it is also true for \(n = k+1\). Hence, by induction, \(u_n = 6^n - 1\) is true for all positive integers.

(b) \(\frac{u_{2n}}{u_n} = \frac{6^{2n} - 1}{6^n - 1} = \frac{(6^n - 1)(6^n + 1)}{6^n - 1} = 6^n + 1\).

Thus, \(u_{2n}\) is divisible by \(u_n\).

(a) The matrix \(\mathbf{M}\) can be decomposed into two transformations: a rotation and a shear. The rotation is represented by \(\begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\), which is an anticlockwise rotation about the origin through an angle \(\theta\). The shear is represented by \(\begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}\), which fixes the x-axis and maps \((0,1)\) to \((2,1)\).

(b) The matrix \(\mathbf{M}\) is given by:

\(\mathbf{M} = \begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}\)

Applying \(\mathbf{M}\) to a point \((x, y)\) gives:

\(\begin{pmatrix} \cos \theta + 2 \sin \theta & 2 \cos \theta - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta \\ x \sin \theta + y \cos \theta \end{pmatrix}\)

\(For a line of invariant points, we require:\)

\(x \sin \theta + y \cos \theta = y\) and \(x \cos \theta + 2x \sin \theta - y \sin \theta + 2y \cos \theta = x\).

Solving these equations gives:

\(x \cos \theta + 2x \sin \theta + \frac{x \sin \theta}{1 - \cos \theta} (2 \cos \theta - \sin \theta) = x\)

Simplifying, we find:

\((\cos \theta + 2 \sin \theta)(1 - \cos \theta) + 2 \sin \theta - \sin^2 \theta = 1 - \cos \theta\)

\(\cos \theta - \cos^2 \theta + 2 \sin^2 \theta - \sin^2 \theta = 1 - \cos \theta \Rightarrow \sin \theta + \cos \theta = 1\)

Thus, \(\theta = \frac{1}{2} \pi\).

(a) The curve is a spiral on \([0, 2\pi]\), starting at the pole with \(\theta = 0\) and increasing as \(\theta\) increases, intersecting the initial line at one other point.

(b) The area \(A\) is given by:

\(A = \frac{1}{2} \int_0^{2\pi} \theta^2 e^{\frac{1}{4}\theta} \, d\theta\)

Integrate by parts twice:

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - \frac{1}{2} \int_0^{2\pi} 8\theta e^{\frac{1}{4}\theta} \, d\theta\)

\(= \frac{1}{2} \left[ 4\theta^2 e^{\frac{1}{4}\theta} \right]_0^{2\pi} - 4 \left[ 4\theta e^{\frac{1}{4}\theta} \right]_0^{2\pi} + 16 \int_0^{2\pi} e^{\frac{1}{4}\theta} \, d\theta\)

\(= \left[ 2\theta^2 e^{\frac{1}{4}\theta} - 16\theta e^{\frac{1}{4}\theta} + 64e^{\frac{1}{4}\theta} \right]_0^{2\pi}\)

\(= (8\pi^2 - 32\pi + 64)e^{\frac{1}{2}\pi} - 64\)

(c) Use \(y = r\sin \theta\) and differentiate:

\(\frac{dy}{d\theta} = \theta e^{\frac{1}{8}\theta} \cos \theta + \sin \theta \left( \frac{1}{8}\theta e^{\frac{1}{8}\theta} + e^{\frac{1}{8}\theta} \right)\)

\(e^{\frac{1}{8}\theta} \neq 0 \Rightarrow \theta \cos \theta + \left( \frac{1}{8}\theta + 1 \right) \sin \theta = 0\)

Check sign change:

\(5\cos 5 + \left( \frac{5}{8} + 1 \right)\sin 5 = -0.1399\)

\(5.05\cos 5.05 + \left( \frac{5.05}{8} + 1 \right)\sin 5.05 = 0.1336\)