Base case: For \(n = 1\),

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

so \(H_1\) is true.

Inductive step: Assume that

\(\sum_{r=1}^{k} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2}.\)

Consider \(\sum_{r=1}^{k+1} r x^{r-1}\):

\(\sum_{r=1}^{k+1} r x^{r-1} = \frac{1 - (k+1)x^k + kx^{k+1}}{(1-x)^2} + (k+1)x^k.\)

Combine over a common denominator:

\(\frac{1 - (k+1)x^k + kx^{k+1} + (k+1)x^k(1-2x+x^2)}{(1-x)^2}.\)

Simplify:

\(\frac{1 + kx^{k+1} + (k+1)x^k(-2x + x^2)}{(1-x)^2} = \frac{1 - (k+2)x^{k+1} + (k+1)x^{k+2}}{(1-x)^2}.\)

Thus, \(H_{k+1}\) is true. By induction, \(H_n\) is true for all positive integers \(n\).

(a) Using Vieta's formulas, we have:

\(b = - (\alpha + \beta + \gamma + \delta) = -3\).

For \(c\), use the formula for the sum of squares:

\(5 = (-3)^2 - 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\).

Solving gives \(c = 2\).

For \(d\), use the given inverse sum:

\(6 = \frac{\alpha\beta\gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta}{\alpha\beta\gamma\delta} = \frac{-d}{-2}\).

Solving gives \(d = 12\).

The equation is \(x^4 - 3x^3 + 2x^2 + 12x - 2 = 0\).

(b) Using the derived quartic equation:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = 3(-27) - 2(5) - 12(3) + 2(4)\).

Calculating gives \(-119\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vectors of the lines: \(\mathbf{d_1} = \begin{pmatrix} -4 \\ 3 \\ 5 \end{pmatrix}\) and \(\mathbf{d_2} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}\).

The vector between a point on \(l_1\) and a point on \(l_2\) is \(\mathbf{b} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

The cross product \(\mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

The shortest distance is given by \(\frac{|\mathbf{b} \cdot (\mathbf{d_1} \times \mathbf{d_2})|}{|\mathbf{d_1} \times \mathbf{d_2}|} = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) The plane \(\Pi\) contains \(l_1\) and the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\). The normal vector to the plane is \(\mathbf{n} = \mathbf{d_1} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -1 \\ -1 \end{pmatrix}\).

The equation of the plane is \(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

(a) Since \(\mathbf{A}\) is singular, its determinant is zero. Calculate the determinant:

\(\det(\mathbf{A}) = 1(1 \cdot 5 - 3 \cdot 2) - k(2 \cdot 5 - 3 \cdot 3) + 3(2 \cdot 2 - 1 \cdot 3) = 0.\)

This simplifies to \(-1 - k + 3 = 0 \Rightarrow k = 2.\)

Substitute \(k = 2\) into \(\mathbf{A}\) and compute \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 \\ 1 & 1 \\ 1 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \\ 1 & 5 & 12 \end{pmatrix}.\)

Then, \(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -11 \\ 1 & 3 & 6 \end{pmatrix} \cdot \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}.\)

(b) For invariant lines, solve \(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = m \begin{pmatrix} x \\ y \end{pmatrix}.\)

This gives \(-9x + 3mx = m(3x - 7mx)\) and \(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9.\)

Thus, \(m = \pm \frac{3}{\sqrt{7}}\), giving lines \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Assume \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF},\) solve:

\(\begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix} - 9 \begin{pmatrix} 0 & \beta \\ 1 & 0 \end{pmatrix}.\)

Equating gives \(\alpha = 3\) and \(\beta = \frac{7}{9}.\)

Thus, \(\mathbf{D} = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}, \mathbf{E} = \begin{pmatrix} \frac{7}{9} & 0 \\ 0 & 1 \end{pmatrix}, \mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.\)

(a) Start with the Cartesian equation \(\left( x - \frac{1}{2} \right)^2 + y^2 = \frac{1}{4}\).

Expand and rearrange: \(x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}\) which simplifies to \(x^2 + y^2 = x\).

In polar coordinates, \(x = r \cos \theta\) and \(x^2 + y^2 = r^2\), so \(r^2 = r \cos \theta\).

Factor to get \(r(r - \cos \theta) = 0\). Since \(r \neq 0\), \(r = \cos \theta\).

(b) Set \(r = \cos \theta = \sin 2\theta\).

Use the identity \(\sin 2\theta = 2 \sin \theta \cos \theta\), so \(\cos \theta = 2 \sin \theta \cos \theta\).

Thus, \(\cos \theta (1 - 2 \sin \theta) = 0\). Since \(\cos \theta \neq 0\), \(\sin \theta = \frac{1}{2}\).

Therefore, \(\theta = \frac{\pi}{6}\) and \(r = \cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\).

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

(c) Sketch the curves \(C_1: r = \cos \theta\) and \(C_2: r = \sin 2\theta\) on the same diagram, marking the point \(P\).

(d) The area \(R\) is given by:

\(\frac{1}{2} \int_0^{\frac{\pi}{6}} \sin^2 2\theta \, d\theta + \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} \cos^2 \theta \, d\theta\).

Use identities: \(\sin^2 2\theta = \frac{1}{2}(1 - \cos 4\theta)\) and \(\cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\).

Integrate to find:

\(\frac{1}{4} \left[ \theta - \frac{1}{4} \sin 4\theta \right]_0^{\frac{\pi}{6}} + \frac{1}{4} \left[ \theta + \frac{1}{2} \sin 2\theta \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}\).

Calculate to get \(\frac{1}{4} \left( \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right) + \frac{1}{4} \left( \frac{\pi}{2} - \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right)\).

Simplify to \(\frac{1}{4} \left( \pi - \sqrt{3} \right)\).