Find _r=1^n (3r + 1). [2 marks]

Evaluate _r=1^10 r^3. [2 marks]

Given (1)/(r) - (1)/(r+2) = (2)/(r(r+2)), what type of method does the sum sum (2)/(r(r+2)) call for? [1 mark]

CCEA CCEA AS 2020-style practice: Find _r=1^n r(r + 2), giving your answer as a fully factorised expression in n.

Expand the term and split using the standard results: _r=1^n r(r + 2) = _r=1^n (r^2 + 2r) = sum r^2 + 2sum r. Substitute sum r^2 = (1)/(6)n(n + 1)(2n + 1) and sum r = (1)/(2)n(n + 1): = (1)/(6)n(n + 1)(2n + 1) + 2 · (1)/(2)n(n + 1) = (1)/(6)n(n + 1)(2n + 1) + n(n + 1). Take out (1)/(6)n(n + 1): = (1)/(6)n(n + 1)[(2n + 1) + 6] = (1)/(6)n(n + 1)(2n + 7). Markers reward splitting the sum, the correct standard results, and full factorisation to (1)/(6)n(n + 1)(2n + 7).

CCEA CCEA AS 2018-style practice: Show that 1r(r + 1) = 1r - 1r + 1, and hence use the method of differences to find _r=1^n 1r(r + 1).

Combine the right-hand side over a common denominator: 1r - 1r + 1 = (r + 1) - rr(r + 1) = 1r(r + 1), as required. Now write the sum out so the middle terms cancel (telescoping): _r=1^n(1r - 1r + 1) = (1 - 12) + (12 - 13) + + (1n - 1n + 1). Only the first and last terms survive: = 1 - 1n + 1 = nn + 1. Markers reward proving the partial fraction, writing out enough terms to show the cancellation, and the closed form (n)/(n + 1).

Northern IrelandFurther MathsSyllabus dot point

How do you sum finite series using standard results and the method of differences?

Summation of finite series using the standard results for the sum of r, r squared and r cubed, and the method of differences for telescoping sums.

A CCEA AS Further Maths answer on summing finite series, the standard results for the sum of r, r squared and r cubed, manipulating sigma notation, and the method of differences for telescoping series.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-16

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking

CCEA wants you to sum finite series using the standard results for $\sum r$ , $\sum r^2$ and $\sum r^3$ , manipulate sigma notation (splitting sums, factoring out constants, changing limits), and apply the method of differences to telescoping series, often after first finding partial fractions.

The answer

The standard results

These are given on the CCEA formula sheet, but you must apply them fluently.

Manipulating sigma notation

After substituting the standard results, always factorise fully, because exam marks are awarded for the simplified form.

The method of differences

The usual route is: split the term into partial fractions, write out the first two or three and the last two terms, identify the cancellation, and quote what survives.

Method of differences with partial fractions

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

step Split into partial fractions

Write $\dfrac{1}{(2r - 1)(2r + 1)} = \dfrac{A}{2r - 1} + \dfrac{B}{2r + 1}$ . Clearing gives $1 = A(2r + 1) + B(2r - 1)$ . Setting $r = \frac{1}{2}$ : $1 = 2A$ , so $A = \frac{1}{2}$ . Setting $r = -\frac{1}{2}$ : $1 = -2B$ , so $B = -\frac{1}{2}$ .

So $\dfrac{1}{(2r - 1)(2r + 1)} = \dfrac{1}{2}\left(\dfrac{1}{2r - 1} - \dfrac{1}{2r + 1}\right).$

step Write out the telescoping sum

\frac{1}{2}\left[\left(\frac{1}{1} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{5}\right) + \cdots + \left(\frac{1}{2n - 1} - \frac{1}{2n + 1}\right)\right].

step Cancel and simplify

Only the first and last terms remain:

\frac{1}{2}\left(1 - \frac{1}{2n + 1}\right) = \frac{1}{2}\cdot\frac{2n}{2n + 1} = \frac{n}{2n + 1}.

Examples in context

Example 1. Total displacement from a varying step. If an object moves $r^2$ units on step $r$ , the total distance after $n$ steps is $\sum r^2 = \frac{1}{6}n(n+1)(2n+1)$ . A closed form lets you predict the position for any $n$ without adding step by step.

Example 2. Convergence in disguise. A telescoping sum such as $\sum \frac{1}{r(r+1)} = \frac{n}{n+1}$ approaches $1$ as $n \to \infty$ . Recognising the telescoping structure is how many infinite series are evaluated exactly rather than estimated.

Try this

Q1. Find $\displaystyle\sum_{r=1}^{n} (3r + 1)$ . [2 marks]

Cue. $3\sum r + \sum 1 = 3 \cdot \frac{1}{2}n(n+1) + n = \frac{1}{2}n(3n + 5)$ .

Q2. Evaluate $\displaystyle\sum_{r=1}^{10} r^3$ . [2 marks]

Cue. $\frac{1}{4}(10)^2(11)^2 = \frac{1}{4}(100)(121) = 3025$ .

Q3. Given $\frac{1}{r} - \frac{1}{r+2} = \frac{2}{r(r+2)}$ , what type of method does the sum $\sum \frac{2}{r(r+2)}$ call for? [1 mark]

Cue. The method of differences (telescoping).

Exam-style practice questions

Practice questions written in the style of CCEA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

CCEA AS 20206 marksFind

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

, giving your answer as a fully factorised expression in

n

Show worked answer →

Expand the term and split using the standard results:

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

Substitute $\sum r^2 = \frac{1}{6}n(n + 1)(2n + 1)$ and $\sum r = \frac{1}{2}n(n + 1)$ :

$= \frac{1}{6}n(n + 1)(2n + 1) + 2 \cdot \frac{1}{2}n(n + 1) = \frac{1}{6}n(n + 1)(2n + 1) + n(n + 1).$

Take out $\frac{1}{6}n(n + 1)$ :

$= \frac{1}{6}n(n + 1)\big[(2n + 1) + 6\big] = \frac{1}{6}n(n + 1)(2n + 7).$

Markers reward splitting the sum, the correct standard results, and full factorisation to $\frac{1}{6}n(n + 1)(2n + 7)$ .

CCEA AS 20187 marksShow that

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

, and hence use the method of differences to find

\displaystyle\sum_{r=1}^{n} \dfrac{1}{r(r + 1)}

Show worked answer →

Combine the right-hand side over a common denominator:

$\dfrac{1}{r} - \dfrac{1}{r + 1} = \dfrac{(r + 1) - r}{r(r + 1)} = \dfrac{1}{r(r + 1)},$ as required.

Now write the sum out so the middle terms cancel (telescoping):

$\sum_{r=1}^{n}\left(\dfrac{1}{r} - \dfrac{1}{r + 1}\right) = \left(1 - \dfrac{1}{2}\right) + \left(\dfrac{1}{2} - \dfrac{1}{3}\right) + \cdots + \left(\dfrac{1}{n} - \dfrac{1}{n + 1}\right).$

Only the first and last terms survive:

$= 1 - \dfrac{1}{n + 1} = \dfrac{n}{n + 1}.$

Markers reward proving the partial fraction, writing out enough terms to show the cancellation, and the closed form $\frac{n}{n + 1}$ .

Related dot points

Sources & how we know this

CCEA GCE Further Mathematics specification — CCEA (2018)