Base case: For \(n = 1\), \(5^{3 \times 1} + 32^1 - 33 = 5^3 + 32 - 33 = 124\), which is divisible by 31.

Inductive step: Assume \(5^{3k} + 32^k - 33\) is divisible by 31 for some positive integer \(k\).

Consider \(5^{3(k+1)} + 32^{k+1} - 33\):

\(5^{3(k+1)} + 32^{k+1} - 33 = 5^{3k+3} + 32^{k+1} - 33\)

\(= (124 + 1)5^{3k} + (31 + 1)32^k - 33\)

\(= 124 \times 5^{3k} + 31 \times 32^k + 5^{3k} + 32^k - 33\)

Since \(124 \times 5^{3k} + 31 \times 32^k\) is divisible by 31, \(5^{3(k+1)} + 32^{k+1} - 33\) is divisible by 31.

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

(a) Use the standard summation formulas:

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

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

Substitute into the expression:

\(6 \left( \frac{1}{6}n(n+1)(2n+1) \right) + 6 \left( \frac{1}{2}n(n+1) \right) - 5n\)

Simplify to get \(2n^3 + 6n^2 - n\).

(b) Divide by the denominator:

\(\frac{6r^2 + 6r - 5}{r^2 + r} = 6 - \frac{5}{r(r+1)}\)

Find partial fractions:

\(\frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}\)

Sum the series:

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

Result: \(6n - 5 + \frac{5}{n+1}\).

(c) Use the result from (b):

\(\sum_{r=n+1}^{2n} \frac{6r^2 + 6r - 5}{r^2 + r} = \sum_{r=1}^{2n} \frac{6r^2 + 6r - 5}{r^2 + r} - \sum_{r=1}^{n} \frac{6r^2 + 6r - 5}{r^2 + r}\)

\(= 12n - 5 + \frac{5}{2n+1} - (6n - 5 + \frac{5}{n+1})\)

Simplify to get \(6n - \frac{5n}{(n+1)(2n+1)}\).

(a) Let \(y = x^2\), so \(x = y^{\frac{1}{2}}\). Substitute into the original equation:

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

Rearrange and square both sides to eliminate the square root:

\((2y^{\frac{1}{2}})^2 = (-y^2 + y - 5)^2\).

Expand and simplify to obtain:

\(y^4 - 2y^3 + 11y^2 - 14y + 25 = 0\).

The value of \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2\) is 2.

(b) Using the relation \(\alpha^2 \beta^2 \gamma^2 \delta^2 = 25\), we have:

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

Substitute the known values to get \(\frac{14}{25}\).

(c) Using the identity:

\(\alpha^4 + \beta^4 + \gamma^4 + \delta^4 = (\alpha^2 + \beta^2 + \gamma^2 + \delta^2)^2 - 2(\alpha^2 \beta^2 + \alpha^2 \gamma^2 + \alpha^2 \delta^2 + \beta^2 \gamma^2 + \beta^2 \delta^2 + \gamma^2 \delta^2)\).

Substitute the known values to get \(2^2 - 2(11) = -18\).

(a) The matrix \(\begin{pmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{pmatrix}\) represents a rotation about the origin through angle \(2\theta\). The transformation sequence is a shear in the x-direction followed by this rotation.

(b) The inverse of \(\mathbf{M}\) is given by \(\mathbf{M}^{-1} = \begin{pmatrix} 1 & -k \\ 0 & 1 \end{pmatrix} \begin{pmatrix} \cos 2\theta & \sin 2\theta \\ -\sin 2\theta & \cos 2\theta \end{pmatrix}\).

(c) To find when \(\mathbf{M} - \mathbf{I}\) is singular, calculate the determinant: \(\det(\mathbf{M} - \mathbf{I}) = (\cos 2\theta - 1)(k \sin 2\theta + \cos 2\theta - 1) - k \sin 2\theta \cos 2\theta + \sin^2 2\theta = 0\). Simplifying gives \(2 - 2\cos 2\theta - k \sin 2\theta = 0\), leading to \(4\sin^2 \theta = 2k \sin \theta \cos \theta\). Using double angle formulas, \(\tan \theta = \frac{1}{2}k\).

(d) Given \(k = 2\sqrt{3}\) and \(\theta = \frac{1}{3}\pi\), the matrix \(\mathbf{M}\) becomes \(\begin{pmatrix} -\frac{1}{2} & -\frac{2\sqrt{3}}{3} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2} \end{pmatrix}\). Transforming \(\begin{pmatrix} x \\ y \end{pmatrix}\) to \(\begin{pmatrix} x \\ y \end{pmatrix}\) gives \(\begin{pmatrix} -\frac{1}{2}x - \frac{2\sqrt{3}}{3}y \\ \frac{\sqrt{3}}{2}x - \frac{1}{2}y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\). Solving gives \(\sqrt{3}x + 3y = 0\).

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

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

Substitute these into the equation: \((r \cos \theta)^2 - (r \sin \theta)^2 = a\).

This simplifies to \(r^2 (\cos^2 \theta - \sin^2 \theta) = a\).

Using the double angle identity, \(\cos 2\theta = \cos^2 \theta - \sin^2 \theta\), we have \(r^2 \cos 2\theta = a\).

Thus, \(r^2 = a \sec 2\theta\).

(b) The sketch shows the curve starting at \(\theta = 0\) and increasing, with the minimum distance from the pole being \(\sqrt{a}\).