1. Base Case: For \(n = 1\), \(7^{2 \times 1} + 97^1 - 50 = 49 + 97 - 50 = 96\), which is divisible by 48.

2. Inductive Hypothesis: Assume \(7^{2k} + 97^k - 50\) is divisible by 48 for some positive integer \(k\).

3. Inductive Step: Consider \(7^{2(k+1)} + 97^{k+1} - 50\).

4. \(7^{2(k+1)} + 97^{k+1} - 50 = 7^{2k+2} + 97^{k+1} - 50 = (48+1)7^{2k} + (96+1)97^k - 50\).

5. This expression is divisible by 48 because \(48 \times 7^{2k} + 96 \times 97^k\) is divisible by 48.

6. Therefore, by induction, \(7^{2n} + 97^n - 50\) is divisible by 48 for all positive integers \(n\).

(a) Expand \((2r+1)^3 - (2r-1)^3\):

\(8r^3 + 12r^2 + 6r + 1 - (8r^3 - 12r^2 + 6r - 1) = 24r^2 + 2\).

Sum both sides and use the method of differences:

\(24 \sum_{r=1}^{n} r^2 + 2n = (2n+1)^3 - 1\).

\(24 \sum_{r=1}^{n} r^2 = 8n^3 + 12n^2 + 4n = 4n(n+1)(2n+1)\).

Thus, \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\).

(b) Express \(S_{2n}\) in terms of sums:

\(S_{2n} = \sum_{r=1}^{2n} r^2 + 2 \sum_{r=1}^{n} (2r)^2 = \sum_{r=1}^{2n} r^2 + 8 \sum_{r=1}^{n} r^2\).

Apply \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\):

\(S_{2n} = \frac{1}{6}(2n)(2n+1)(4n+1) + \frac{8}{6}n(n+1)(2n+1)\).

\(S_{2n} = \frac{1}{3}n(2n+1)(4n+4+4) = \frac{1}{3}n(2n+1)(8n+5)\).

(c) Evaluate the limit:

\(\lim_{n \to \infty} \frac{S_{2n}}{n^3} = \frac{16}{3}\).

(a) To find the Cartesian equation of the plane \(\Pi\), we first find a normal vector to the plane by taking the cross product of the direction vectors of the lines:

Direction vectors: \(\begin{pmatrix} -1 \\ 2 \\ 1 \end{pmatrix}\) and \(\begin{pmatrix} 3 \\ 2 \\ -1 \end{pmatrix}\).

Cross product: \(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -1 & 2 & 1 \\ 3 & 2 & -1 \end{vmatrix} = \begin{pmatrix} -4 \\ 2 \\ -8 \end{pmatrix}\).

Normal vector: \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Using point \((3, -2, 1)\) on the plane, substitute into \(2x - y + 4z = d\):

\(2(3) - (-2) + 4(1) = 12\).

Thus, the equation is \(2x - y + 4z = 12\).

(b) The line \(l\) is parallel to \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}\). The normal to the plane is \(\begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix}\).

Dot product: \(\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \sqrt{2} \sqrt{21} \cos \alpha\).

\(\cos \alpha = \frac{3}{\sqrt{42}}\).

Acute angle: \(90^\circ - \alpha = 27.6^\circ\).

(c) The position vector of the foot of the perpendicular \(\overrightarrow{OF}\) is given by:

\(\overrightarrow{OF} = \overrightarrow{OP} + t \begin{pmatrix} 2 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} 2 + 2t \\ 3 - t \\ 1 + 4t \end{pmatrix}\).

Substitute into the plane equation: \(2(2 + 2t) - (3 - t) + 4(1 + 4t) = 12\).

Solve for \(t\): \(2t + 5 = 12\), \(t = \frac{7}{2}\).

\(\overrightarrow{OF} = \frac{1}{3} \begin{pmatrix} 8 \\ 8 \\ 7 \end{pmatrix}\).

(a) The matrix \(M\) is a product of two matrices: \(\begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}\) represents a shear in the x-direction, and \(\begin{pmatrix} \frac{1}{2}\sqrt{2} & -\frac{1}{2}\sqrt{2} \\ \frac{1}{2} & \frac{1}{2}\sqrt{2} \end{pmatrix}\) represents a rotation anticlockwise about the origin through 45°.

(b) To find \(M^{-1}\), we calculate the inverse of the product of the two matrices: \(M^{-1} = \begin{pmatrix} \frac{k+1}{\sqrt{2}} & \frac{k-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} \end{pmatrix}\).

(c) For no invariant lines through the origin, the characteristic equation of \(M\) must have no real roots. The characteristic equation is derived from \(\det(M - \lambda I) = 0\). Solving gives \(k^2 + 4k - 4 < 0\), leading to the solution \(2(-1-\sqrt{2}) < k < 2(-1+\sqrt{2})\).

(a) Start with the Cartesian equation \((x^2 + y^2)^2 = 36(x^2 - y^2)\).

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

Substitute: \(r^4 = 36(r^2 \cos^2 \theta - r^2 \sin^2 \theta)\).

Factor out \(r^2\): \(r^4 = 36r^2(\cos^2 \theta - \sin^2 \theta)\).

Use the identity \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\): \(r^2 = 36 \cos 2\theta\).

(b) The sketch is a closed curve symmetrical about the initial line. The maximum distance from the pole is 6.

(c) The area enclosed by \(C\) is given by:

\(\frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} r^2 \, d\theta = \frac{1}{2} \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} 36 \cos 2\theta \, d\theta\).

\(= 18 \int_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi} \cos 2\theta \, d\theta\).

\(= 18 [\frac{1}{2} \sin 2\theta]_{-\frac{1}{4}\pi}^{\frac{1}{4}\pi}\).

\(= 18 [\sin \frac{1}{2}\pi - \sin(-\frac{1}{2}\pi)]\).

\(= 18 [1 - (-1)] = 18 \times 2 = 18\).

(d) To find the maximum distance from the initial line, set \(y = 6 \cos^{\frac{1}{2}} 2\theta \sin \theta\).

Set \(\frac{dy}{d\theta} = 0\) and solve:

\(\cos^2 2\theta \cos \theta - \cos^{\frac{1}{2}} 2\theta \sin 2\theta \sin \theta = 0\).

Use trigonometric identities to solve for \(\theta\):

\(\theta = \pm \frac{1}{4}\pi\).

The maximum distance is \(3\sqrt{2}\).