(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

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

(b) Express \(\frac{1}{(2-3r)(5-3r)}\) as partial fractions: \(\frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Write the sum: \(\sum_{r=1}^{n} \frac{1}{3} \left( \frac{1}{2-3r} - \frac{1}{5-3r} \right)\).

Observe cancellation: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum converges to \(-\frac{1}{6}\).

(a) Expand \((2-3r)(5-3r) = 9r^2 - 21r + 10\).

Use standard summation formulas: \(\sum r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum r = \frac{1}{2}n(n+1)\).

Substitute: \(9\left(\frac{1}{6}n(n+1)(2n+1)\right) - 21\left(\frac{1}{2}n(n+1)\right) + 10n\).

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

(b) Partial fraction decomposition: \(\frac{1}{(2-3r)(5-3r)} = \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Sum: \(\sum_{r=1}^{n} \frac{1}{3}\left(\frac{1}{2-3r} - \frac{1}{5-3r}\right)\).

Telescoping series: \(-\frac{1}{6} + \frac{1}{3(2-3n)}\).

(c) As \(n \to \infty\), the term \(\frac{1}{3(2-3n)} \to 0\).

Thus, the sum is \(\frac{1}{6}\).

(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) We start with \(\sum_{r=1}^{n} (2r+1) = 2\sum_{r=1}^{n} r + n\). Using the formula for the sum of the first \(n\) natural numbers, \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\), we have:

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

(b) To show \(\frac{2r+1}{(r^2+1)(r^2+2r+2)} = \frac{1}{r^2+1} - \frac{1}{r^2+2r+2}\), we find a common denominator:

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

(c) Using the result from (b), the series becomes a telescoping series:

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

(d) As \(n \to \infty\), the term \(\frac{1}{n^2+2n+2} \to 0\), so the sum converges to \(\frac{1}{2}\).

(a) Start by expressing \(\frac{5k}{(5r+k)(5r+5+k)}\) using partial fractions:

\(\frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5r+k} - \frac{1}{5r+5+k} \right)\)

Then, the sum becomes:

\(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5+k} + \frac{1}{10+k} + \frac{1}{15+k} + \ldots + \frac{1}{5n+k} - \frac{1}{5n+5+k} \right)\)

This simplifies to:

\(= k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

(b) Given \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\), we have:

\(\frac{k}{5+k} = \frac{1}{3}\)

Solving for \(k\):

\(3k = 5 + k \Rightarrow 2k = 5 \Rightarrow k = \frac{5}{2}\)

(c) Using the result from (a), find:

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

\(= k \left( \frac{1}{5+k} - \frac{1}{5n^2+5+k} \right) - k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

\(= k \left( \frac{1}{5n+5+k} - \frac{1}{5n^2+5+k} \right)\)

Substitute \(k = \frac{5}{2}\):

\(= \frac{1}{2n+1} - \frac{1}{2n^2+3}\)

(a) Start by expressing \(\frac{5k}{(5r+k)(5r+5+k)}\) using partial fractions:

\(\frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5r+k} - \frac{1}{5r+5+k} \right)\)

Then, the sum becomes:

\(\sum_{r=1}^{n} \frac{5k}{(5r+k)(5r+5+k)} = k \left( \frac{1}{5+k} + \frac{1}{10+k} + \frac{1}{15+k} + \ldots + \frac{1}{5n+k} - \frac{1}{5+k} - \frac{1}{10+k} - \ldots - \frac{1}{5n+5+k} \right)\)

By the method of differences, this simplifies to:

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

(b) Given \(\sum_{r=1}^{\infty} \frac{5k}{(5r+k)(5r+5+k)} = \frac{1}{3}\), we have:

\(k \left( \frac{1}{5+k} \right) = \frac{1}{3}\)

Solving for \(k\):

\(\frac{k}{5+k} = \frac{1}{3}\)

\(3k = 5 + k\)

\(2k = 5\)

\(k = \frac{5}{2}\)

(c) Using the result from (a), find:

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

\(= k \left( \frac{1}{5+k} - \frac{1}{5n^2+5+k} \right) - k \left( \frac{1}{5+k} - \frac{1}{5n+5+k} \right)\)

\(= k \left( \frac{1}{5n+5+k} - \frac{1}{5n^2+5+k} \right)\)

Substitute \(k = \frac{5}{2}\):

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

\(= \frac{1}{2n+1} - \frac{1}{2n^2+3}\)

(a) Expand \(r(r+1)(3r+4)\) to get \(3r^3 + 7r^2 + 4r\).

Use standard summation formulae:

\(\sum_{r=1}^{N} r^3 = \left( \frac{N(N+1)}{2} \right)^2, \quad \sum_{r=1}^{N} r^2 = \frac{N(N+1)(2N+1)}{6}, \quad \sum_{r=1}^{N} r = \frac{N(N+1)}{2}.\)

Substitute these into the expanded expression:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Simplify to get:

\(\frac{1}{12}N(9N^3 + 46N^2 + 75N + 38) = \frac{1}{12}N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Substitute into the series:

\(\sum_{r=1}^{N} \left( \frac{4}{r} - \frac{1}{r+1} \right) \left( \frac{1}{4} \right)^{r+1}.\)

Write out terms and simplify using the method of differences:

\(\left( \frac{1}{4^2} \right) \left( 4 - \frac{1}{2} \right) + \left( \frac{1}{4^3} \right) \left( 4 - \frac{1}{3} \right) + \ldots + \left( \frac{1}{4^{N+1}} \right) \left( 4 - \frac{1}{N+1} \right).\)

The series telescopes to:

\(\frac{1}{4} \left( 1 - \frac{1}{N+1} \right).\)

(c) As \(N \to \infty\), the series becomes:

\(\frac{1}{4}.\)

(a) Expand the expression \(\sum_{r=1}^{N} r(r+1)(3r+4)\) as follows:

\(3 \sum_{r=1}^{N} r^3 + 7 \sum_{r=1}^{N} r^2 + 4 \sum_{r=1}^{N} r.\)

Using standard formulae:

\(\frac{3}{4} N^2 (N+1)^2 + \frac{7}{6} N(N+1)(2N+1) + 2N(N+1).\)

Simplify to:

\(\frac{1}{12} N (9N^3 + 46N^2 + 75N + 38) = \frac{1}{12} N(N+1)(N+2)(9N+19).\)

(b) Express \(\frac{3r+4}{r(r+1)}\) in partial fractions:

\(\frac{3r+4}{r(r+1)} = \frac{4}{r} - \frac{1}{r+1}.\)

Then,

\(\sum_{r=1}^{N} \frac{3r+4}{r(r+1)} \left( \frac{1}{4} \right)^{r+1} = \left( \frac{1}{4^2} \right) \left( 4 - 1 \right) + \left( \frac{1}{4^3} \right) \left( 4 - \frac{1}{2} \right) + \ldots + \left( \frac{1}{4^{N+1}} \right) \left( 4 - \frac{1}{N+1} \right).\)

This simplifies to:

\(\frac{1}{4} - \frac{1}{4^{N+1}(N+1)}.\)

(c) As \(N \to \infty\), the term \(\frac{1}{4^{N+1}(N+1)} \to 0\), so the sum converges to:

\(\frac{1}{4}\).

(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides:

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

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

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

(b) \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an\)

\(\frac{1}{2} n(n+1) + an = n\)

Solving for \(a\):

\(a = \frac{1}{2}(1-n)\)

(a) Use the standard results for summation: \(\sum_{r=1}^{n} r = \frac{n(n+1)}{2}\) and \(\sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}\).

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

Substitute the standard results: \(3 \left( \frac{n(n+1)(2n+1)}{6} \right) + 3 \left( \frac{n(n+1)}{2} \right) + n\).

Simplify to get \(\frac{1}{2} n(n+1)(2n+1) + \frac{3}{2} n(n+1) + n = n^3 + 3n^2 + 3n\).

(b) Start with \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{(r+1)^3 - r^3}{r^3 (r+1)^3}\).

Expand \((r+1)^3 - r^3 = r^3 + 3r^2 + 3r + 1 - r^3 = 3r^2 + 3r + 1\).

Thus, \(\frac{1}{r^3} - \frac{1}{(r+1)^3} = \frac{3r^2 + 3r + 1}{r^3 (r+1)^3}\).

Use the method of differences: \(\sum_{r=1}^{n} \left( \frac{1}{r^3} - \frac{1}{(r+1)^3} \right) = 1 - \frac{1}{(n+1)^3}\).

(c) As \(n \to \infty\), \(\frac{1}{(n+1)^3} \to 0\), so the sum converges to 1.

(a) Let \(y = 3x + 1\), then \(x = \frac{1}{3}(y - 1)\).

Substitute into the original equation:

\(27\left(\frac{y-1}{3}\right)^3 + 18\left(\frac{y-1}{3}\right)^2 + 6\left(\frac{y-1}{3}\right) - 1 = 0\)

\(\Rightarrow (y-1)^3 + 2(y-1)^2 + 2(y-1) - 1 = 0\)

\(\Rightarrow y^3 - 3y^2 + 3y - 1 + 2y^2 - 4y + 2 + 2y - 2 - 1 = 0\)

\(\Rightarrow y^3 - y^2 + y - 2 = 0\)

(b) \(S_2 = 1^2 - 2(1) = -1\)

\(S_3 = (3\alpha + 1)^3 + (3\beta + 1)^3 + (3\gamma + 1)^3 = -1 - (1) + 6 = 4\)

(c) \(S_{-1} = \frac{(3\alpha + 1)(3\beta + 1) + (3\beta + 1)(3\gamma + 1) + (3\gamma + 1)(3\alpha + 1)}{(3\alpha + 1)(3\beta + 1)(3\gamma + 1)} = \frac{1}{2}\)

\(2S_{-2} = S_1 + S_{-1} = 1 - 3 + \frac{1}{2}\)

\(S_{-2} = -\frac{3}{4}\)

(a) Consider \((r+1)^2 - r^2 = r^2 + 2r + 1 - r^2 = 2r + 1\).

Using the method of differences, sum both sides from \(r=1\) to \(n\):

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

This simplifies to:

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

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

(b) Given \(\sum_{r=1}^{n} (r+a) = \sum_{r=1}^{n} r + an = n\).

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

\(\frac{1}{2} n(n+1) + an = n.\)

Solving for \(a\):

\(an = n - \frac{1}{2} n(n+1)\)

\(a = \frac{1}{2} (1-n).\)

(a) Using the identity for the sum of squares of roots: \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

From the cubic equation, \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\).

Thus, \(\alpha^2 + \beta^2 + \gamma^2 = (-4)^2 - 2(6) = 16 - 12 = 4\).

(b) Expand \((\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2\).

Therefore, \(\sum_{r=1}^{n} ((\alpha + r)^2 + (\beta + r)^2 + (\gamma + r)^2) = \sum_{r=1}^{n} (\alpha^2 + 2\alpha r + r^2 + \beta^2 + 2\beta r + r^2 + \gamma^2 + 2\gamma r + r^2)\).

Using \(\alpha + \beta + \gamma = -4\) and \(\alpha^2 + \beta^2 + \gamma^2 = 4\), we have:

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

Using standard summation formulas:

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

Thus, \(3 \sum_{r=1}^{n} r^2 - 8 \sum_{r=1}^{n} r + 4n = 3 \cdot \frac{n(n+1)(2n+1)}{6} - 8 \cdot \frac{n(n+1)}{2} + 4n\).

Simplifying gives:

\(\frac{n(n+1)(2n+1)}{2} - 4n(n+1) + 4n\).

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

Thus, \(a = -\frac{5}{2}, b = \frac{1}{2}\).

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

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Now, apply the method of differences:

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

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite series, the terms after the first cancel out, leaving:

\(\frac{1}{k}\)

(c) Use the same method of differences for the range \(r=n\) to \(r=n^2\):

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

This results in:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(a) Using the formula for the sum of squares of roots, \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

Given \(\alpha + \beta + \gamma = -4\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 6\), we have:

\((-4)^2 - 2(6) = 16 - 12 = 4\).

(b) Expand \((\alpha + r)^2 = \alpha^2 + 2\alpha r + r^2\).

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^2 + (\beta + r)^2 + (\gamma + r)^2) = \sum_{r=1}^{n} (4 + 2(-4)r + 3r^2)\).

Calculate: \(4n - 4n(n+1) + \frac{1}{2}n(n+1)(2n+1)\).

Simplify: \(-4n^2 + \frac{1}{2}n(n+1)(2n+1)\).

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

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

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

\(\frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k} \left( \frac{1}{kr-k+1} - \frac{1}{kr+1} \right)\)

Thus, the sum becomes:

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

This simplifies to:

\(\frac{1}{k} \left( 1 - \frac{1}{kn+1} \right)\)

(b) For the infinite sum, as \(n \to \infty\), the term \(\frac{1}{kn+1} \to 0\), so:

\(\sum_{r=1}^{\infty} \frac{1}{(kr+1)(kr-k+1)} = \frac{1}{k}\)

(c) Using the result from part (a), the sum from \(n\) to \(n^2\) is:

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

Using the formula from part (a):

\(= \frac{1}{k} \left( 1 - \frac{1}{kn^2+1} \right) - \frac{1}{k} \left( 1 - \frac{1}{k(n-1)+1} \right)\)

This simplifies to:

\(\frac{1}{k} \left( \frac{1}{k(n-1)+1} - \frac{1}{kn^2+1} \right)\)

(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) 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) 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) We start with \(\sum_{r=1}^{n} r(r+2) = \sum_{r=1}^{n} (r^2 + 2r)\).

Using the formulae, \(\sum_{r=1}^{n} r^2 = \frac{1}{6}n(n+1)(2n+1)\) and \(\sum_{r=1}^{n} r = \frac{1}{2}n(n+1)\).

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

Simplifying gives \(\frac{1}{6}n(n+1)(2n+7)\).

(b) Express \(\frac{1}{r(r+2)}\) in partial fractions: \(\frac{1}{r(r+2)} = \frac{1}{2} \left( \frac{1}{r} - \frac{1}{r+2} \right)\).

Then, \(\sum_{r=1}^{n} \frac{1}{r(r+2)} = \frac{1}{2} \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots + \frac{1}{n} - \frac{1}{n+2} \right)\).

This simplifies to \(\frac{1}{2} \left( \frac{3}{2} - \frac{1}{n+1} - \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the terms \(\frac{1}{n+1}\) and \(\frac{1}{n+2}\) approach zero.

Thus, the sum converges to \(\frac{1}{2} \times \frac{3}{2} = \frac{3}{4}\).

(a) Use the standard summation formulas:

\(\sum_{r=1}^{n} 1 = n\)

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

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

Substitute these into the expression:

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

Simplify to get:

\(\frac{1}{3}n - n^2 - \frac{1}{3}n^3\)

(b) Start with the given expression:

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

Verify by expanding and simplifying both sides to show equality.

Use the method of differences:

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

This simplifies to a telescoping series:

\(= \frac{2}{5} - \frac{1}{2} + \frac{3}{10} - \frac{2}{5} + \frac{4}{17} - \frac{3}{10} + \ldots + \frac{n+1}{(n+1)^2 + 1} - \frac{n}{n^2 + 1}\)

Resulting in:

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

(a) We start with \(\sum_{r=1}^{n} r(r+1)(r+2) = \sum_{r=1}^{n} (r^3 + 3r^2 + 2r)\).

Using standard summation formulae:

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

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

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

Substituting these into the expression:

\(\frac{1}{4}n(n+1)^2 + \frac{1}{2}n(n+1)(2n+1) + n(n+1)\)

Factorising gives \(\frac{1}{4}n(n+1)(n+2)(n+3)\).

(b) Express \(\frac{1}{r(r+1)(r+2)}\) in partial fractions:

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

Using the method of differences:

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

This telescopes to:

\(\frac{1}{2}f(1) - f(2) + \frac{1}{2}f(3) + \frac{1}{2}f(2) - f(3) + \frac{1}{2}f(4) + \ldots + \frac{1}{2}f(n) - f(n+1) + \frac{1}{2}f(n+2)\)

Resulting in \(\frac{1}{4} \left( \frac{1}{n+1} + \frac{1}{n+2} \right)\).

(c) As \(n \to \infty\), the sum becomes \(\frac{1}{4}\).

(a) Using the formula for the sum of squares of roots, \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\).

From the cubic equation, \(\alpha + \beta + \gamma = -2\) and \(\alpha\beta + \beta\gamma + \gamma\alpha = 3\).

Thus, \((-2)^2 - 2(3) = 4 - 6 = -2\).

(b) Using the formula for the sum of cubes of roots, \(\alpha^3 + \beta^3 + \gamma^3 = 3(\alpha\beta\gamma) + (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha)\).

\(\alpha\beta\gamma = -3\), so \(\alpha^3 + \beta^3 + \gamma^3 = 3(-3) + (-2)(-2 - 3) = -9 + 10 = 1\).

(c) Expanding \((\alpha + r)^3 = \alpha^3 + 3\alpha^2 r + 3\alpha r^2 + r^3\).

Thus, \(\sum_{r=1}^{n} ((\alpha + r)^3 + (\beta + r)^3 + (\gamma + r)^3) = \sum_{r=1}^{n} (1 + 3(-2)r + 3(-2)r^2 + 3r^3)\).

Using standard results, \(n - 6(\frac{1}{2}n(n+1)) - 6(\frac{1}{6}n(n+1)(2n+1)) + \frac{3}{4}n^2(n+1)^2\).

Simplifying gives \(n + \frac{1}{4}n(n+1)(-12 - 4(2n+1) + 3n(n+1))\).

Finally, \(n + \frac{1}{4}n(n+1)(3n^2 - 5n - 16)\).

Answer:

1. \(a=\frac{49}{3}\), \(b=56\), \(c=\frac{143}{3}\), so \(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+56n^2+\frac{143}{3}n\).
2. \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\).
3. \(\sum_{r=1}^{\infty}\frac{1}{(7r+1)(7r+8)}=\frac{1}{56}\).

1. Expand the product inside the sum:

   \((7r+1)(7r+8)=49r^2+56r+7r+8=49r^2+63r+8\).

   So

   \(\sum_{r=1}^{n}(7r+1)(7r+8)=\sum_{r=1}^{n}(49r^2+63r+8)\).

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

   \(\sum_{r=1}^{n}(49r^2+63r+8)=49\left(\frac{1}{6}n(n+1)(2n+1)\right)+63\left(\frac{1}{2}n(n+1)\right)+8n\).

   Expanding gives

   \(49\left(\frac{1}{6}n(n+1)(2n+1)\right)=\frac{49}{3}n^3+\frac{49}{2}n^2+\frac{49}{6}n\),

   and

   \(63\left(\frac{1}{2}n(n+1)\right)=\frac{63}{2}n^2+\frac{63}{2}n\).

   Therefore

   \(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+\left(\frac{49}{2}+\frac{63}{2}\right)n^2+\left(\frac{49}{6}+\frac{63}{2}+8\right)n\).

   Hence

   \(\sum_{r=1}^{n}(7r+1)(7r+8)=\frac{49}{3}n^3+56n^2+\frac{143}{3}n\).

   So \(a=\frac{49}{3}\), \(b=56\), and \(c=\frac{143}{3}\).

2. First write the fraction in partial fractions:

   \(\frac{1}{(7r+1)(7r+8)}=\frac{A}{7r+1}+\frac{B}{7r+8}\).

   Multiplying through by \((7r+1)(7r+8)\),

   \(1=A(7r+8)+B(7r+1)\).

   Putting \(r=-\frac{1}{7}\) gives \(1=7A\), so \(A=\frac{1}{7}\). Putting \(r=-\frac{8}{7}\) gives \(1=-7B\), so \(B=-\frac{1}{7}\).

   Thus

   \(\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{7r+1}-\frac{1}{7r+8}\right)\).

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

   \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\sum_{r=1}^{n}\left(\frac{1}{7r+1}-\frac{1}{7r+8}\right)\).

   Writing out the terms,

   \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{15}+\frac{1}{15}-\frac{1}{22}+\frac{1}{22}-\frac{1}{29}+\cdots+\frac{1}{7n+1}-\frac{1}{7n+8}\right)\).

   The intermediate terms cancel, leaving

   \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\).

3. From part (b),

   \(\sum_{r=1}^{n}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\left(\frac{1}{8}-\frac{1}{7n+8}\right)\).

   As \(n\to\infty\), \(\frac{1}{7n+8}\to 0\). Therefore

   \(\sum_{r=1}^{\infty}\frac{1}{(7r+1)(7r+8)}=\frac{1}{7}\cdot\frac{1}{8}=\frac{1}{56}\).

Answer:

1. (a) \(a=2\) and \(b=1\), so \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac{3}{4}-\frac{2n+1}{2n(n+1)}\).
2. (b) \(\sum_{r=2}^{\infty}\frac{1}{r^{2}-1}=\frac{3}{4}\).
3. (c) The limit is \(\frac{1}{2}\).

1. Factorise the denominator: \(r^{2}-1=(r-1)(r+1)\). Write

   \(\frac{1}{r^{2}-1}=\frac{1}{(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r+1}\).

   Multiplying by \((r-1)(r+1)\) gives \(1=A(r+1)+B(r-1)\). Substituting \(r=1\) gives \(A=\frac12\), and substituting \(r=-1\) gives \(B=-\frac12\). Hence

   \(\frac{1}{r^{2}-1}=\frac12\left(\frac{1}{r-1}-\frac{1}{r+1}\right)\).

   Therefore

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac12\left(\left(1-\frac13\right)+\left(\frac12-\frac14\right)+\left(\frac13-\frac15\right)+\cdots+\left(\frac{1}{n-1}-\frac{1}{n+1}\right)\right)\).

   Most terms cancel, leaving

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac12\left(1+\frac12-\frac1n-\frac{1}{n+1}\right)\).

   So

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac12\left(\frac1n+\frac{1}{n+1}\right)=\frac34-\frac{2n+1}{2n(n+1)}\).

   Comparing this with \(\frac34-\frac{an+b}{2n(n+1)}\), we get \(a=2\) and \(b=1\).

2. From part (a),

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac{2n+1}{2n(n+1)}\).

   As \(n\to\infty\), \(\frac{2n+1}{2n(n+1)}\to 0\). Therefore

   \(\sum_{r=2}^{\infty}\frac{1}{r^{2}-1}=\frac34\).

3. Write the required finite sum as a difference of two sums:

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

   Using part (a) with upper limit \(2n\),

   \(\sum_{r=2}^{2n}\frac{1}{r^{2}-1}=\frac34-\frac{4n+1}{4n(2n+1)}\).

   Also,

   \(\sum_{r=2}^{n}\frac{1}{r^{2}-1}=\frac34-\frac{2n+1}{2n(n+1)}\).

   Hence

   \(\sum_{r=n+1}^{2n}\frac{n}{r^{2}-1}=n\left(\frac{2n+1}{2n(n+1)}-\frac{4n+1}{4n(2n+1)}\right)\).

   So

   \(\sum_{r=n+1}^{2n}\frac{n}{r^{2}-1}=\frac{2n+1}{2(n+1)}-\frac{4n+1}{4(2n+1)}\).

   Taking the limit as \(n\to\infty\),

   \(\frac{2n+1}{2(n+1)}\to 1\), while \(\frac{4n+1}{4(2n+1)}\to \frac12\). Therefore the required limit is

   \(1-\frac12=\frac12\).

Answer: (i) \(\displaystyle \sum_{r=1}^{N}\frac{1}{(3r+1)(3r-2)}=\frac13-\frac{1}{3(3N+1)}\).

(ii) \(\displaystyle \lim_{N\to\infty}\sum_{r=N+1}^{N^2}\frac{N}{(3r+1)(3r-2)}=\frac19\).

(i) First use partial fractions:

\(\dfrac{1}{(3r+1)(3r-2)}=\dfrac13\left(\dfrac{1}{3r-2}-\dfrac{1}{3r+1}\right)\).

Therefore

\(\displaystyle \sum_{r=1}^{N}\dfrac{1}{(3r+1)(3r-2)}=\dfrac13\sum_{r=1}^{N}\left(\dfrac{1}{3r-2}-\dfrac{1}{3r+1}\right)\).

Writing out the terms,

\(\displaystyle \dfrac13\left(1-\dfrac14+\dfrac14-\dfrac17+\cdots+\dfrac{1}{3N-2}-\dfrac{1}{3N+1}\right)\).

All the middle terms cancel, leaving

\(\displaystyle \dfrac13\left(1-\dfrac{1}{3N+1}\right)=\dfrac13-\dfrac{1}{3(3N+1)}\),

as required.

(ii) Use the result from part (i) twice:

\(\displaystyle \sum_{r=N+1}^{N^2}\frac{N}{(3r+1)(3r-2)}=N\left(\sum_{r=1}^{N^2}\frac{1}{(3r+1)(3r-2)}-\sum_{r=1}^{N}\frac{1}{(3r+1)(3r-2)}\right)\).

So

\(\displaystyle =N\left(\frac13-\frac{1}{3(3N^2+1)}-\frac13+\frac{1}{3(3N+1)}\right)\)

\(\displaystyle =\frac{N}{3(3N+1)}-\frac{N}{3(3N^2+1)}\).

Putting over a common denominator gives

\(\displaystyle \frac{N^3-N^2}{(3N+1)(3N^2+1)}\).

Hence, as \(N\to\infty\), the limit is

\(\displaystyle \frac{1}{9}\).

Answer: (i) \(S_N=\frac{1}{3}N(25N^2+90N+83)\).

(ii) \(T_N=\frac{1}{30}\left(\frac{1}{6}-\frac{1}{5N+6}\right)=\frac{N}{30(5N+6)}\).

(iii) \(\displaystyle \lim_{N\to\infty} \left(N^{-3}S_NT_N\right)=\frac{5}{18}\).

We work with the two sums separately.

(i) Expand the general term of \(S_N\):

\((5r+1)(5r+6)=25r^2+35r+6.\)

So

\(S_N=\sum_{r=1}^N (25r^2+35r+6)=25\sum_{r=1}^N r^2+35\sum_{r=1}^N r+6\sum_{r=1}^N 1.\)

Using the standard results

\(\sum_{r=1}^N r=\frac{N(N+1)}{2},\qquad \sum_{r=1}^N r^2=\frac{N(N+1)(2N+1)}{6},\qquad \sum_{r=1}^N 1=N,\)

we get

\(S_N=25\cdot \frac{N(N+1)(2N+1)}{6}+35\cdot \frac{N(N+1)}{2}+6N.\)

Put everything over 6:

\(S_N=\frac{25N(N+1)(2N+1)+105N(N+1)+36N}{6}.\)

Now expand:

\(25N(N+1)(2N+1)=25N(2N^2+3N+1)=50N^3+75N^2+25N,\)

so

\(S_N=\frac{50N^3+75N^2+25N+105N^2+105N+36N}{6}\)

\(=\frac{50N^3+180N^2+166N}{6}=\frac13 N(25N^2+90N+83).\)

This is the required result.

(ii) First rewrite the summand by partial fractions:

\(\frac{1}{(5r+1)(5r+6)}=\frac{A}{5r+1}+\frac{B}{5r+6}.\)

Equivalently, since the denominator differs by 5, a convenient choice is

\(\frac{1}{(5r+1)(5r+6)}=\frac{1}{5}\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right),\)

because

\(\frac{1}{5}\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right)=\frac{1}{5}\cdot \frac{5}{(5r+1)(5r+6)}.\)

Hence

\(T_N=\frac15\sum_{r=1}^N\left(\frac{1}{5r+1}-\frac{1}{5r+6}\right).\)

Now write out the first few terms:

\(T_N=\frac15\left(\frac16-\frac1{11}+\frac1{11}-\frac1{16}+\cdots+\frac{1}{5N+1}-\frac{1}{5N+6}\right).\)

Everything cancels except the first positive term and the last negative term, so

\(T_N=\frac15\left(\frac16-\frac{1}{5N+6}\right).\)

This can also be simplified to

\(T_N=\frac{1}{30}-\frac{1}{5(5N+6)}=\frac{N}{30(5N+6)}.\)

(iii) From part (i),

\(S_N=\frac13N(25N^2+90N+83).\)

From part (ii),

\(T_N=\frac{N}{30(5N+6)}.\)

Therefore

\(N^{-3}S_NT_N=N^{-3}\cdot \frac13N(25N^2+90N+83)\cdot \frac{N}{30(5N+6)}.\)

Simplify the powers of \(N\):

\(N^{-3}S_NT_N=\frac{25N^2+90N+83}{90(5N+6)N}.\)

Divide numerator and denominator by \(N\):

\(N^{-3}S_NT_N=\frac{25N+90+83/N}{90(5+6/N)}.\)

As \(N\to\infty\), this tends to

\(\frac{25}{90\cdot 5}=\frac{5}{18}.\)

So the limit is \(\frac{5}{18}\).

Answer: (i) \(S_{2n}=-n(2n+1)\)

(ii) \(\displaystyle \lim_{n\to\infty}\frac{S_{2n}}{n^2}=-2\), and \(\displaystyle \lim_{n\to\infty}\frac{S_{2n+1}}{n^2}=2\).

We are given \(S_n=\sum_{r=1}^n (-1)^{r-1}r^2\).

(i) Consider \(S_{2n}\):

\(S_{2n}=1^2-2^2+3^2-4^2+\cdots+(2n-1)^2-(2n)^2\).

Group the terms in pairs:

\(S_{2n}=(1^2-2^2)+(3^2-4^2)+\cdots+((2n-1)^2-(2n)^2)\).

For the \(k\)th pair,

\((2k-1)^2-(2k)^2=(4k^2-4k+1)-4k^2=1-4k\).

So

\(S_{2n}=\sum_{k=1}^n (1-4k)=n-4\sum_{k=1}^n k\).

Using \(\sum_{k=1}^n k=\frac{n(n+1)}{2}\),

\(S_{2n}=n-4\cdot\frac{n(n+1)}{2}=n-2n(n+1)=n-2n^2-2n=-2n^2-n\).

Hence

\(S_{2n}=-n(2n+1)\).

(ii) From part (i),

\(\frac{S_{2n}}{n^2}=\frac{-n(2n+1)}{n^2}=-\left(2+\frac1n\right)\).

Therefore

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

Now find \(S_{2n+1}\). Since

\(S_{2n+1}=S_{2n}+(2n+1)^2\),

we have

\(S_{2n+1}=-n(2n+1)+(2n+1)^2\).

Factor out \((2n+1)\):

\(S_{2n+1}=(2n+1)(-n+2n+1)=(2n+1)(n+1)\).

So

\(\frac{S_{2n+1}}{n^2}=\frac{(2n+1)(n+1)}{n^2}=(2+\frac1n)(1+\frac1n)\).

Taking the limit gives

\(\lim_{n\to\infty}\frac{S_{2n+1}}{n^2}=2\).

Answer: EITHER: the three roots written down are \(1,e^{2\pi i/3},e^{-2\pi i/3}\). The other three roots are \(-1,\frac{1}{3}+\frac{2\sqrt2}{3}i,\frac{1}{3}-\frac{2\sqrt2}{3}i\).

OR: one possible choice is \(\mathbf P=\begin{pmatrix}-2&0&0\\1&3&0\\0&4&1\end{pmatrix}\) and \(\mathbf D=\begin{pmatrix}3^n&0&0\\0&7^n&0\\0&0&1\end{pmatrix}\). The final range is \(-\frac{1}{7}\lt k\lt\frac{1}{7}\).

EITHER

(i) Since \(z=e^{i\theta}\), write \(z=e^{i\theta/2}e^{i\theta/2}\). Then

\(\frac{z-1}{z+1}=\frac{e^{i\theta/2}-e^{-i\theta/2}}{e^{i\theta/2}+e^{-i\theta/2}}.\)

Using \(e^{iu}-e^{-iu}=2i\sin u\) and \(e^{iu}+e^{-iu}=2\cos u\),

\(\frac{z-1}{z+1}=\frac{2i\sin(\theta/2)}{2\cos(\theta/2)}=i\tan\frac{\theta}{2}.\)

(ii) If \(z=1\), then each numerator is zero, so the sum is zero.

For the other cube roots of unity, write \(z=e^{i\theta}\), where \(\theta=\pm\frac{2\pi}{3}\). Since \(z^3=1\), the first term is zero. Also, by part (i),

\(\frac{z^2-1}{z^2+1}=i\tan\theta,\qquad \frac{z-1}{z+1}=i\tan\frac{\theta}{2}.\)

If \(\theta=\frac{2\pi}{3}\), then \(\tan\theta=-\sqrt3\) and \(\tan\frac{\theta}{2}=\sqrt3\). If \(\theta=-\frac{2\pi}{3}\), the two values are reversed in sign. In both cases, their sum is zero. Therefore

\(\frac{z^3-1}{z^3+1}+\frac{z^2-1}{z^2+1}+\frac{z-1}{z+1}=0.\)

(iii) From part (ii), the three cube roots of unity are roots of the polynomial equation. Thus

\(z=1,\quad z=e^{2\pi i/3},\quad z=e^{-2\pi i/3}\)

are three roots.

Expanding the left-hand side gives

\(3z^6+z^5+z^4-z^2-z-3=0.\)

This factorises as

\(3z^6+z^5+z^4-z^2-z-3=(z+1)(z^3-1)(3z^2-2z+3).\)

The roots not already listed are therefore

\(z=-1\)

and the roots of \(3z^2-2z+3=0\). Hence

\(z=\frac{2\pm\sqrt{4-36}}{6}=\frac{2\pm i\sqrt{32}}{6}=\frac{1}{3}\pm\frac{2\sqrt2}{3}i.\)

So the other three roots are

\(\boxed{-1,\ \frac{1}{3}+\frac{2\sqrt2}{3}i,\ \frac{1}{3}-\frac{2\sqrt2}{3}i}.\)

OR

(i) Another eigenvector corresponding to \(\lambda\) is any non-zero scalar multiple of \(\mathbf e\), for example \(2\mathbf e\).

(ii) Since \(\mathbf A\mathbf e=\lambda\mathbf e\), repeated application gives

\(\mathbf A^n\mathbf e=\lambda^n\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf A^n\), with eigenvalue \(\lambda^n\).

(iii) The matrix \(\mathbf A\) is triangular, so its eigenvalues are \(3,7,1\).

For \(\lambda=3\), an eigenvector is \((-2,1,0)^T\).

For \(\lambda=7\), an eigenvector is \((0,3,4)^T\).

For \(\lambda=1\), an eigenvector is \((0,0,1)^T\).

Therefore one suitable matrix is

\(\mathbf P=\begin{pmatrix}-2&0&0\\1&3&0\\0&4&1\end{pmatrix}.\)

For \(\mathbf A^n\), the diagonal matrix is

\(\mathbf D=\begin{pmatrix}3^n&0&0\\0&7^n&0\\0&0&1\end{pmatrix}.\)

Thus

\(\mathbf A^n=\mathbf P\mathbf D\mathbf P^{-1}.\)

(iv) The partial sum is

\(\sum_{n=1}^{N}k^n(\mathbf A^n-k\mathbf A^{n+1})=k\mathbf A-k^{N+1}\mathbf A^{N+1}.\)

For the infinite sum to equal \(k\mathbf A\), we need

\(k^{N+1}\mathbf A^{N+1}\to \mathbf 0\)

as \(N\to\infty\). Since the largest eigenvalue of \(\mathbf A\) in modulus is \(7\), this requires

\(|7k|\lt1.\)

Hence

\(\boxed{-\frac{1}{7}\lt k\lt\frac{1}{7}}.\)

Answer: (i) \(\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}=\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\).

(ii) \(S_N=\dfrac{1}{\mathrm e}-\dfrac{1}{(N+1)\mathrm e^{N+1}}\), so \(S=\dfrac{1}{\mathrm e}\).

(iii) The least value is \(N=6\).

For the identity, put the right-hand side over the common denominator \(n(n+1)\mathrm e^{n+1}\):

\(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}=\dfrac{(n+1)\mathrm e-n}{n(n+1)\mathrm e^{n+1}}=\dfrac{n(\mathrm e-1)+\mathrm e}{n(n+1)\mathrm e^{n+1}}.\)

Therefore

\(S_N=\sum_{n=1}^{N}\left(\dfrac{1}{n\mathrm e^n}-\dfrac{1}{(n+1)\mathrm e^{n+1}}\right).\)

This is a telescoping sum:

\(S_N=\left(\dfrac1{\mathrm e}-\dfrac1{2\mathrm e^2}\right)+\left(\dfrac1{2\mathrm e^2}-\dfrac1{3\mathrm e^3}\right)+\cdots+\left(\dfrac1{N\mathrm e^N}-\dfrac1{(N+1)\mathrm e^{N+1}}\right).\)

All intermediate terms cancel, so

\(S_N=\dfrac1{\mathrm e}-\dfrac1{(N+1)\mathrm e^{N+1}}.\)

Hence

\(S=\lim_{N\to\infty}S_N=\dfrac1{\mathrm e}.\)

Then

\(S-S_N=\dfrac1{(N+1)\mathrm e^{N+1}}\),

so

\((N+1)(S-S_N)=\dfrac1{\mathrm e^{N+1}}.\)

We need \(\mathrm e^{-(N+1)}\lt 10^{-3}\). Since \(\ln 1000\approx 6.91\), we need \(N+1>6.91\), so the least integer value is

\(N=6\).