Past Exam Questions

(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 \(a\):

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