Base case: For \(n = 1\), \(u_1 = 4 = 3^1 + 1\). This holds true.

Inductive step: Assume it is true for \(n = k\), so \(u_k = 3^k + 1\).

Then \(u_{k+1} = 3u_k - 2 = 3(3^k + 1) - 2 = 3^{k+1} + 3 - 2 = 3^{k+1} + 1\).

Thus, it is true for \(n = k+1\). Hence, by induction, \(u_n = 3^n + 1\) for all positive integers \(n\).

(a) To find the equation of the plane \(\Pi\), we need a normal vector to the plane. The direction vector of \(l_1\) is \(\mathbf{i} - \mathbf{j} - 4\mathbf{k}\), and the plane is parallel to \(2\mathbf{i} + 5\mathbf{j} - 4\mathbf{k}\). The normal vector is the cross product of these two vectors:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -1 & -4 \\ 2 & 5 & -4 \end{vmatrix} = \mathbf{i}(16 - 20) - \mathbf{j}(-4 - (-8)) + \mathbf{k}(5 + 2) = -4\mathbf{i} + 4\mathbf{j} + 7\mathbf{k}\)

Using the point \((1, 3, -1)\) on \(l_1\), the equation of the plane is:

\(24(1) - 4(3) + 7(-1) = d \Rightarrow 24x - 4y + 7z = 5\).

(b) To find the acute angle between \(l_2\) and \(\Pi\), we use the dot product of the direction vector of \(l_2\), \(5\mathbf{i} - 5\mathbf{j} - 2\mathbf{k}\), and the normal vector of \(\Pi\), \(24\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\):

\((24, -4, 7) \cdot (5, -5, -2) = 24 \times 5 + (-4) \times (-5) + 7 \times (-2) = 120 + 20 - 14 = 126\).

The magnitudes are \(\sqrt{24^2 + (-4)^2 + 7^2} = \sqrt{641}\) and \(\sqrt{5^2 + (-5)^2 + (-2)^2} = \sqrt{54}\).

\(\cos \alpha = \frac{126}{\sqrt{641} \times \sqrt{54}}\).

The angle \(\alpha\) is the angle between the line and the normal, so the acute angle between \(l_2\) and \(\Pi\) is \(90^\circ - \alpha = 42.6^\circ\).

(a) We use the identity \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Substituting the given values, \(3 = 2^2 - 2(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta)\).

Solving, \(\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = \frac{1}{2}\).

(b) We use the identity \(\alpha^2(\beta + \gamma + \delta) + \beta^2(\alpha + \gamma + \delta) + \gamma^2(\alpha + \beta + \delta) + \delta^2(\alpha + \beta + \gamma)\).

This simplifies to \(\alpha^2(2-\alpha) + \beta^2(2-\beta) + \gamma^2(2-\gamma) + \delta^2(2-\delta)\).

Substituting the given values, \(2(3) - 4 = 2\).

(c)(i) Using the polynomial equation, \(6S_4 - 12S_3 + 3S_2 + 2S_1 + 24 = 0\).

Substitute \(S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_4 = \frac{11}{6}\).

(c)(ii) Using the polynomial equation, \(6S_5 - 12S_4 + 3S_3 + 2S_2 + S_1 = 0\).

Substitute \(S_4 = \frac{11}{6}, S_3 = 4, S_2 = 3, S_1 = 2\) to find \(S_5 = -\frac{4}{3}\).

(a) To find \(CAB\), first calculate \(AB\):

\(AB = \begin{pmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 2 & 5 \end{pmatrix} \begin{pmatrix} 0 & -2 \\ -1 & 3 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} -2 & 4 \\ -1 & 3 \\ -4 & 7 \end{pmatrix}\)

Then calculate \(CAB\):

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

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

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

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

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

Thus, \(y = \frac{3}{7}x\) and \(y = -\frac{3}{7}x\).

(c) The matrix \(M\) represents a stretch parallel to the x-axis with scale factor 3.

(d) Find \(N\) such that \(NM = CAB\):

\(M^{-1} = \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix}\)

\(N = CAB \cdot M^{-1} = \begin{pmatrix} 3 & -7 \\ -9 & 3 \end{pmatrix} \begin{pmatrix} \frac{1}{3} & 0 \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -7 \\ -3 & 3 \end{pmatrix}\)

(a) The series \(S_n = \sum_{r=1}^{n} (x^{f(r)} - x^{f(r+1)})\) telescopes to \(x^{f(1)} - x^{f(n+1)}\).

(b) With \(f(r) = \ln r\), the series becomes \(\sum_{r=1}^{\infty} (x^{\ln r} - x^{\ln(r+1)})\). For convergence, \(0 < x \leq 1\). If \(x < 1\), the series converges to \(1\). If \(x = 1\), the series converges to \(0\).

(c) With \(f(r) = 2 \log_x r\), \(x^{2 \log_x r} = r^2\). Thus, \(S_n = 1 - (n+1)^2 = -n^2 - 2n\). Using standard results, \(\sum_{n=1}^{N} (-n^2 - 2n) = -\frac{1}{6} N(N+1)(2N+7)\).