Base case: For \(n = 1\), \(2025^1 + 47^1 - 2 = 2070 = 45 \times 46\), which is divisible by 46.

Inductive step: Assume \(2025^k + 47^k - 2\) is divisible by 46 for some positive integer \(k\).

Then, consider \(2025^{k+1} + 47^{k+1} - 2\).

\(2025^{k+1} + 47^{k+1} - 2 = (2024 + 1)2025^k + (46 + 1)47^k - 2\)

This can be rewritten as:

\(2024 \times 2025^k + 46 \times 47^k + 2025^k + 47^k - 2\)

\(= 46(44 \times 2025^k + 47^k) + (2025^k + 47^k - 2)\)

By the inductive hypothesis, \(2025^k + 47^k - 2\) is divisible by 46.

Hence, \(2025^{k+1} + 47^{k+1} - 2\) is divisible by 46.

Therefore, by induction, the statement is true for every positive integer \(n\).

(a) Using the identity \((\alpha + \beta + \gamma + \delta)^2 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\), we have:

\(0 = \alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(7)\)

Thus, \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = -14\).

(b) Using the given equation, \(S_4 + 7S_2 + 3S_1 + 88 = 0\), where \(S_4 = \alpha^4 + \beta^4 + \gamma^4 + \delta^4\), we find:

\(S_4 = 10\).

(c) Expanding \((r + \alpha^2)^2 = r^2 + 2\alpha^2 r + \alpha^4\), we have:

\(\sum_{r=1}^{10} ((r + \alpha^2)^2 + (r + \beta^2)^2 + (r + \gamma^2)^2 + (r + \delta^2)^2) = \sum_{r=1}^{10} (4r^2 - 28r + 10)\)

\(= 4\left(\frac{10}{6}(10+1)(2(10)+1)\right) - 14(10)(10+1) + 10(10)\)

\(= 100\).

(a) Consider \(w_{r+1} = (r+1)(r+2)\ldots(r+10)\) and \(w_r = r(r+1)(r+2)\ldots(r+9)\).

Then, \(w_{r+1} - w_r = (r+1)(r+2)\ldots(r+10) - r(r+1)(r+2)\ldots(r+9)\).

Factor out \((r+1)(r+2)\ldots(r+9)\):

\(= (r+1)(r+2)\ldots(r+9)((r+10) - r) = 10(r+1)(r+2)\ldots(r+9)\).

(b) The sum \(\sum_{r=1}^{n} u_r = \frac{1}{10}((w_2 - w_1) + (w_3 - w_2) + \ldots + (w_{n+1} - w_n))\).

This telescopes to \(\frac{1}{10}(w_{n+1} - w_1)\).

\(w_{n+1} = (n+1)(n+2)\ldots(n+10)\) and \(w_1 = 1 \cdot 2 \cdot \ldots \cdot 10 = 10!\).

Thus, \(\sum_{r=1}^{n} u_r = \frac{1}{10}((n+1)(n+2)\ldots(n+10) - 10!)\).

(c) The series \(v_1 + v_2 + \ldots + v_n = x^{w_2} - x^{w_1} + x^{w_3} - x^{w_2} + \ldots + x^{w_{n+1}} - x^{w_n}\).

This telescopes to \(x^{w_{n+1}} - x^{w_1}\).

For convergence as \(n \to \infty\), \(|x| < 1\) is required.

Thus, for \(-1 < x < 1\), the sum to infinity is \(-x^{w_1} = -x^{10!}\).

For \(x = \pm 1\), the sum to infinity is 0.

(a) To find the Cartesian equation of the plane \(\Pi\), we need a normal vector. The direction vectors are \(\mathbf{i} - 2\mathbf{j} - \mathbf{k}\) and \(3\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\). The normal vector is the cross product:

\(\begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -2 & -1 \\ 3 & 2 & -2 \end{vmatrix} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

Using the point \((2, 3, -2)\) on the plane, substitute into \(6x - y + 8z = d\):

\(6(2) - 3 + 8(-2) = -7\)

Thus, the equation is \(6x - y + 8z = -7\).

(b) The position vector of the foot of the perpendicular \(\mathbf{F}\) is given by:

\(\mathbf{OF} = \mathbf{OP} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\mathbf{OF} = \begin{pmatrix} 4 \\ 2 \\ 9 \end{pmatrix} + t \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix} = \begin{pmatrix} 4 + 6t \\ 2 - t \\ 9 + 8t \end{pmatrix}\)

Substitute into the plane equation:

\(6(4 + 6t) - (2 - t) + 8(9 + 8t) = -7\)

\(101t + 94 = -7\)

\(t = -1\)

\(\mathbf{OF} = \begin{pmatrix} -2 \\ 3 \\ 1 \end{pmatrix}\)

(c) The acute angle \(\alpha\) between \(l\) and \(\Pi\) is found using:

\(\cos \alpha = \frac{\mathbf{d} \cdot \mathbf{n}}{\|\mathbf{d}\| \|\mathbf{n}\|}\)

\(\mathbf{d} = \begin{pmatrix} 3 \\ 5 \\ -1 \end{pmatrix}, \mathbf{n} = \begin{pmatrix} 6 \\ -1 \\ 8 \end{pmatrix}\)

\(\cos \alpha = \frac{5}{\sqrt{35} \sqrt{101}}\)

\(\alpha = 4.8^\circ\)

(a) The vertical asymptote occurs where the denominator is zero: \(x + a = 0 \Rightarrow x = -a\). For the oblique asymptote, divide \(x^2 + a\) by \(x + a\) to get \(y = x - a\).

(b) Differentiate \(y = \frac{x^2 + a}{x + a}\) to find stationary points: \(\frac{dy}{dx} = \frac{(x+a)^2 - (x^2+a)}{(x+a)^2} = \frac{-a^2}{(x+a)^2}\). Set \(\frac{dy}{dx} = 0\) to find \(x = -a \pm \sqrt{a^2 + a}\).

(c) Sketch the curve showing the asymptotes \(x = -a\) and \(y = x - a\). The curve intersects the y-axis at \((0, 1)\).

(d) Sketch the absolute value of the curve from (c), reflecting any negative parts above the x-axis.

(e) Solve \(\left| \frac{x^2 + a}{x + a} \right| = a\). This leads to \(x^2 - ax + a - a^2 = 0\). The discriminant \(a^2 - 4(a-a^2) = 5a^2 - 4a > 0\) gives \(a > \frac{3}{4}\).