(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides:

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

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

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

(b) \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an\)

\(\frac{1}{2} n(n+1) + an = n\)

Solving for \(a\):

\(a = \frac{1}{2}(1-n)\)

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

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

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

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

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

So \(H_{k+1}\) 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 equation and given \(\alpha^3 + \beta^3 + \gamma^3 + \delta^3 = -27\):

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

Calculating gives \(-119\).

(a) To find the shortest distance between \(l_1\) and \(l_2\), we first find the direction vector from one line to another: \(\begin{pmatrix} 2 \\ -2 \\ 3 \end{pmatrix} - \begin{pmatrix} -2 \\ -3 \\ -5 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix}\).

Next, find the common perpendicular by taking the cross product of the direction vectors of \(l_1\) and \(l_2\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & 3 & 5 \\ 2 & -3 & 1 \end{vmatrix} = \begin{pmatrix} 18 \\ 14 \\ 6 \end{pmatrix} \sim \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix}\).

Use the formula for shortest distance:

\(\frac{1}{\sqrt{139}} \left| \begin{pmatrix} 4 \\ 1 \\ 8 \end{pmatrix} \cdot \begin{pmatrix} 9 \\ 7 \\ 3 \end{pmatrix} \right| = \frac{67}{\sqrt{139}} \approx 5.68\).

(b) To find the equation of the plane \(\Pi\), find a vector perpendicular to the plane using the cross product of the direction vector of \(l_1\) and the vector from the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) to a point on \(l_1\):

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ -4 & 3 & 5 \end{vmatrix} = \begin{pmatrix} 3 \\ -3 \\ -1 \end{pmatrix} \sim \begin{pmatrix} 9 \\ -3 \\ -1 \end{pmatrix}\).

Use the point \(-\mathbf{i} - 3\mathbf{j} - 4\mathbf{k}\) in the plane equation:

\(1(-1) + 3(-3) - 1(-4) = -6 \Rightarrow x + 3y - z = -6\).

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

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

Solving gives \(k = 2\).

Calculate \(\mathbf{CAB}\):

\(\mathbf{C} \cdot \mathbf{A} = \begin{pmatrix} -2 & -1 & 1 \\ 1 & 1 & 3 \end{pmatrix} \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix}\)

\(\mathbf{CAB} = \begin{pmatrix} -2 & -4 & -4 \\ 12 & 9 & 18 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix}\)

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

\(-9x + 3mx = m(3x - 7mx)\)

\(-9 + 3m = 3m - 7m^2 \Rightarrow 7m^2 = 9\)

\(m = \pm \frac{3}{\sqrt{7}}\)

Invariant lines: \(y = \frac{3}{\sqrt{7}}x\) and \(y = -\frac{3}{\sqrt{7}}x\).

(c) Given \(\mathbf{CAB} = \mathbf{D} - 9\mathbf{EF}\), set \(\mathbf{D} = \begin{pmatrix} \alpha & 0 \\ 0 & \alpha \end{pmatrix}\), \(\mathbf{E} = \begin{pmatrix} \beta & 0 \\ 0 & 1 \end{pmatrix}\), \(\mathbf{F} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\).

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

\(\alpha = 3, \beta = \frac{7}{9}\)

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