(a) Consider \(\sum_{r=1}^{n} e^{rx}(e^{2x} - 2e^x + 1) = e^x - e^{2x} - e^{(n+1)x} + e^{(n+2)x}\).

Using the method of differences, the terms cancel out in a pattern, leading to the final result.

(b) For convergence, \(e^{nx} \to 0\) as \(n \to \infty\), which occurs when \(x < 0\). The sum to infinity is \(e^x - e^{2x}\).

(c) Using the laws of logarithms, \(\sum_{r=1}^{n} \ln u_r = \sum_{r=1}^{n} (rx + \ln(e^x - 1)^2)\).

Applying the formula \(\sum_{r=1}^{n} r = \frac{1}{2} n(n+1)\), we get \(\frac{1}{2} xn(n+1) + n \ln(e^x - 1)^2\).

(a) The matrix \(\mathbf{A} = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\) represents a shear in the \(x\)-direction.

(b) Base case: For \(n = 1\), \(\mathbf{A}^1 = \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix}\), which is true.

Inductive step: Assume true for \(n = k\), so \(\mathbf{A}^k = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix}\).

Then \(\mathbf{A}^{k+1} = \mathbf{A}^k \mathbf{A} = \begin{pmatrix} 1 & ka \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & a \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & a + ak \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & (k+1)a \\ 0 & 1 \end{pmatrix}\).

Thus, by induction, \(\mathbf{A}^n = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix}\) for all positive integers \(n\).

(c) \(\mathbf{A}^n \mathbf{B} = \begin{pmatrix} 1 & na \\ 0 & 1 \end{pmatrix} \begin{pmatrix} b & b \\ a^{-1} & a^{-1} \end{pmatrix} = \begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix}\).

Transform \(\begin{pmatrix} x \\ y \end{pmatrix}\) by multiplying matrices to find \(\begin{pmatrix} X \\ Y \end{pmatrix}\).

\(\begin{pmatrix} b+n & b+n \\ \frac{1}{a} & \frac{1}{a} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} (b+n)(x+y) \\ \frac{1}{a}(x+y) \end{pmatrix}\).

\(\frac{1}{a}(1+m) = m(b+n)(1+m)\)

\(y = -x\)

\(y = \frac{1}{a(b+n)}x\)

(a) Start with the polar coordinates: \(x = r \cos \theta\) and \(y = r \sin \theta\). Substitute into the Cartesian equation:

\(x^2 + xy + y^2 = r^2 \cos^2 \theta + r^2 \cos \theta \sin \theta + r^2 \sin^2 \theta = a\).

This simplifies to \(r^2 (\cos^2 \theta + \sin^2 \theta + \cos \theta \sin \theta) = a\).

Using \(\cos^2 \theta + \sin^2 \theta = 1\) and \(\sin 2\theta = 2 \sin \theta \cos \theta\), we have:

\(r^2 (1 + \frac{1}{2} \sin 2\theta) = a\).

Thus, \(r^2 = \frac{2a}{2 + \sin 2\theta}\).

(b) The sketch should show a polar graph with the curve in the correct domain \(0 \leq \theta \leq \frac{1}{4}\pi\), with \(r\) strictly decreasing.

(c) Given \(\sin 2\theta = \frac{2 \tan \theta}{1 + \tan^2 \theta}\), substitute \(t = \tan \theta\), then \(\frac{d\theta}{dt} = \frac{1}{1 + t^2}\).

The integral becomes:

\(R = \frac{1}{2} a \int_{0}^{1} \frac{1}{t^2 + t + 1} \, dt\).

Complete the square: \(t^2 + t + 1 = (t + \frac{1}{2})^2 + \frac{3}{4}\).

Integrate: \(R = \frac{1}{\sqrt{3}} [\arctan(\frac{2t + 1}{\sqrt{3}})]_{0}^{1}\).

Evaluate: \(R = \frac{\pi a}{6\sqrt{3}}\).

(a) The direction vector of line AB is \(\mathbf{AB} = 4\mathbf{i} - \mathbf{j} + \mathbf{k}\).

The direction vector of line CD is \(\mathbf{CD} = \mathbf{j} + (\lambda - 3)\mathbf{k}\).

Find the common perpendicular using the cross product and the formula for perpendicular distance:

\(\frac{1}{\sqrt{(\lambda - 2)^2 + (4\lambda - 12)^2 + 16}} \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ 0 & 1 & \lambda - 3 \end{vmatrix} = 3\)

Solving gives \(\lambda^2 - 5\lambda + 4 = 0\).

(b)(i) The equation of \(\Pi_1\) is \(\mathbf{r} = 7\mathbf{i} + 4\mathbf{j} - \mathbf{k} + s(4\mathbf{i} - \mathbf{j} + \mathbf{k}) + t(-5\mathbf{i} + 3\mathbf{j} + 2\mathbf{k})\).

(b)(ii) The normal to \(\Pi_2\) is found using the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 4 & -1 & 1 \\ -5 & 3 & 7 \end{vmatrix} = -8\mathbf{i} - 25\mathbf{j} + 7\mathbf{k}\)

Substitute a point to find \(-8x - 25y + 7z = -163\).

(c) The angle between \(\Pi_1\) and \(\Pi_2\) is found using the dot product of normals:

\(\cos \theta = \frac{414}{\sqrt{243 \times 738}}\)

\(\theta = 12.1^\circ\).

(a) From the roots \(\alpha, \alpha, \gamma\), we have:

\(2\alpha + \gamma = -b\)

\(\alpha^2 + 2\alpha\gamma = 0\)

Solving, \(\alpha = -2\gamma\) leading to \(-4\gamma + \gamma = -b\), so \(\gamma = \frac{1}{3}b\) and \(\alpha = -\frac{2}{3}b\).

\(\alpha^2\gamma = -d\) leading to \(\frac{4}{27}b^3 = -d\) leading to \(4b^3 + 27d = 0\).

(b) Given \(2\alpha^2 + \gamma^2 = 3b\), substitute for \(\alpha\) and \(\gamma\):

\(3b = b^2\) leading to \(b = 3\).

Substitute back to find \(d\):

\(d = -4\).