(a) Start by expressing \(\frac{1}{(ar+1)(ar+a+1)}\) using partial fractions:

\(\frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{ar+1} - \frac{1}{ar+a+1} \right)\)

Now, sum from \(r=1\) to \(n\):

\(\sum_{r=1}^{n} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} + \frac{1}{2a+1} + \frac{1}{3a+1} + \ldots + \frac{1}{an+a+1} \right)\)

Using the method of differences, this simplifies to:

\(\frac{1}{a} \left( \frac{1}{a+1} - \frac{1}{an+a+1} \right)\)

(b) For the infinite series:

\(\sum_{r=1}^{\infty} \frac{1}{(ar+1)(ar+a+1)} = \frac{1}{a} \left( \frac{1}{a+1} \right) = \frac{1}{6}\)

This leads to:

\(\frac{1}{a^2 + a} = \frac{1}{6}\)

Solving \(a^2 + a - 6 = 0\) gives \(a = 2\).

(a) To find the equation of the plane, we first determine the direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\):

\(\overrightarrow{AB} = (-4 - 4)\mathbf{i} + (3 + 4)\mathbf{j} + (-4 - 1)\mathbf{k} = -8\mathbf{i} + 7\mathbf{j} - 5\mathbf{k}\)

\(\overrightarrow{AC} = (4 - 4)\mathbf{i} + (-1 + 4)\mathbf{j} + (-2 - 1)\mathbf{k} = 0\mathbf{i} + 3\mathbf{j} - 3\mathbf{k}\)

The normal vector \(\mathbf{n}\) to the plane is given by the cross product \(\overrightarrow{AB} \times \overrightarrow{AC}\):

\(\mathbf{n} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 7 & -5 \\ 0 & 3 & -3 \end{vmatrix} = \mathbf{i}(7(-3) - (-5)(3)) - \mathbf{j}(-8(-3) - (-5)(0)) + \mathbf{k}(-8(3) - 7(0))\)

\(= \mathbf{i}(-21 + 15) - \mathbf{j}(24) + \mathbf{k}(-24) = -6\mathbf{i} - 24\mathbf{j} - 24\mathbf{k}\)

Normalizing gives \(\mathbf{n} = \mathbf{i} + 4\mathbf{j} + 4\mathbf{k}\).

Using point \(A(4, -4, 1)\), the equation is:

\(x + 4y + 4z = -8\).

(b) The perpendicular distance from the origin \(O(0, 0, 0)\) to the plane is:

\(\frac{|0 + 4(0) + 4(0) + 8|}{\sqrt{1^2 + 4^2 + 4^2}} = \frac{8}{\sqrt{33}} \approx 1.39\).

(c) The line \(OD\) has the equation \(\mathbf{r} = t(2\mathbf{i} + 3\mathbf{j} - 3\mathbf{k})\).

Substitute into the plane equation:

\(2t + 12t - 12t = -8\)

\(2t = -8\)

\(t = -4\)

Coordinates of intersection: \((-8, -12, 12)\).

(a) Base case: \(u_1 > 4\) (given).

Inductive step: Assume \(u_k > 4\) for some positive integer \(k\).

Then \(u_{k+1} - 4 = \frac{u_k^2 + u_k + 12}{2u_k} - 4 = \frac{u_k^2 - 7u_k + 12}{2u_k}\).

\(u_{k+1} - 4 = \frac{(u_k - 3)(u_k - 4)}{2u_k} > 0\).

Hence, by induction, \(u_n > 4\) for all positive integers \(n\).

(b) \(u_{n+1} - u_n = \frac{-u_n^2 + u_n + 12}{2u_n} = \frac{-(u_n - 4)(u_n + 3)}{2u_n}\).

So \(u_n > 4 \Rightarrow u_{n+1} - u_n < 0\).

(a) Let \(y = x^{-3}\), then \(x = y^{-\frac{1}{3}}\). Substitute into the original equation:

\(2y^{-1} + 5y^{-\frac{2}{3}} - 6 = 0\)

Multiply through by \(y\) to eliminate the fractions:

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(5y^{-\frac{2}{3}} = 6 - 2y^{-1}\)

\(125y = (6y - 2)^3\)

Expanding and simplifying gives:

\(216y^3 - 216y^2 - 53y - 8 = 0\)

(b) Using the identity \(\alpha^{-3} + \beta^{-3} + \gamma^{-3} = 1\) and \(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3} = -\frac{53}{216}\), we find:

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = (\alpha^{-3} + \beta^{-3} + \gamma^{-3})^2 - 2(\alpha^{-3}\beta^{-3} + \beta^{-3}\gamma^{-3} + \gamma^{-3}\alpha^{-3})\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = 1^2 - 2\left(-\frac{53}{216}\right)\)

\(\alpha^{-6} + \beta^{-6} + \gamma^{-6} = \frac{161}{108}\)

(c) Using the equation \(216S_9 = 216S_6 + 53S_3 + 24\) and substituting the values from parts (a) and (b):

\(S_9 = \frac{399}{216} = \frac{133}{72}\)

(a) To find vertical asymptotes, set the denominator equal to zero: \(x^2 + x + 1 = 0\). The discriminant is \(b^2 - 4ac = 1^2 - 4 \times 1 \times 1 = -3\), which is less than zero, so no real roots exist. Therefore, there are no vertical asymptotes. The horizontal asymptote is found by considering the degrees of the polynomials. As \(x \to \infty\), \(y \to \frac{2x^2}{x^2} = 2\). Thus, the horizontal asymptote is \(y = 2\).

(b) To find stationary points, differentiate \(y\) with respect to \(x\):

\(\frac{dy}{dx} = \frac{(x^2 + x + 1)(4x - 1) - (2x^2 - x - 1)(2x + 1)}{(x^2 + x + 1)^2}\).

Set \(\frac{dy}{dx} = 0\) to find critical points:

\(3x^2 + 6x = 0\).

Solving gives \(x = 0\) or \(x = -2\). Substituting back into the original equation gives points \((0, -1)\) and \((-2, 3)\).

(c) The curve intersects the x-axis where \(y = 0\):

\(2x^2 - x - 1 = 0\).

Solving gives \(x = 1\) and \(x = -\frac{1}{2}\). The curve intersects the y-axis where \(x = 0\), giving \((0, -1)\).

(d) For \(\left| \frac{2x^2 - x - 1}{x^2 + x + 1} \right| = k\) to have 4 distinct real solutions, the graph must intersect the line \(y = k\) at 4 points. This occurs when \(0 < k < 1\).