Summation of Series | 9231 Further Pure P2

What is Summation?

A series is the sum of the terms of a sequence. In 9231 Paper 2, you are expected to find closed-form expressions for sums like:

General finite sum S = Σ f(r) for r = 1 to n We seek a formula in n that gives the exact total without adding term by term.

The Cambridge 9231 syllabus tests three distinct techniques. You must recognise which applies — sometimes two are needed in combination.

Technique 1

Standard Results
Memorise the formulae for Σr, Σr², Σr³ and use them directly or in combination.

Technique 2

Method of Differences
Express f(r) = g(r) − g(r−1). The sum telescopes and nearly all terms cancel.

Technique 3

Partial Fractions
Split a rational f(r) into partial fractions first, then apply Method of Differences.

Sigma Notation — a Quick Reminder

Make sure you are fluent with the following sigma algebra before attempting series questions:

Linearity Σ[af(r) + bg(r)] = a·Σf(r) + b·Σg(r)

Constant term Σ c = cn (where c is constant)

Partial sum Σ(r=m to n) f(r) = Σ(r=1 to n) f(r) − Σ(r=1 to m−1) f(r) Used when the lower limit is not 1 — very common in Cambridge questions.

Index shift Σ(r=1 to n) f(r+1) = Σ(r=2 to n+1) f(r) Shift when a substitution changes the summation variable.

⚠️ Common Pitfall — Lower Limit ≠ 1

Cambridge often asks for Σ from r = 3 to n, or r = 2 to 2n. Always subtract the missing initial terms. Writing the full formula and subtracting is safer than guessing at a new formula.

🎯 Paper 2 Exam Pattern

A typical question first asks you to show a partial fraction or difference form — this is the setup for part (ii).
Part (ii) then asks you to sum to n terms using the result in part (i).
Part (iii) often asks for the sum from r = k to n, or for the sum to infinity (only if the series converges).
Always simplify your final answer — Cambridge mark schemes expect a neat closed form.

The Standard Results

These four results must be memorised. They are not provided in the Cambridge formula booklet for P2.

Series	Closed Form	Degree in n	Memory Hint
Σ 1 (r=1 to n)	n	Linear	Trivial — n ones
Σ r	½n(n+1)	Quadratic	"Triangular numbers" — Gauss's formula
Σ r²	⅙n(n+1)(2n+1)	Cubic	Three factors: n, n+1, 2n+1 divided by 6
Σ r³	[½n(n+1)]²	Quartic	(Σr)² — beautiful symmetry

💡 Why Σr³ = (Σr)²?

This is one of the most elegant results in elementary number theory. The sum of the first n cubes equals the square of the sum of the first n integers. Verify: 1³+2³+3³ = 36 = 6² = (1+2+3)². It holds for all n — try proving it by induction (Lesson 2.2)!

Using Standard Results in Combination

Most exam questions require you to expand f(r) as a polynomial, then apply the results term-by-term.

E

Using Standard Results — Combination

Find Σ r(2r − 1) for r = 1 to n.

Step 1

Expand: r(2r − 1) = 2r² − r

Step 2

Apply linearity: Σ(2r² − r) = 2·Σr² − Σr

Step 3

Substitute standard results:
= 2 · ⅙n(n+1)(2n+1) − ½n(n+1)
= ⅓n(n+1)(2n+1) − ½n(n+1)

Step 4

Factorise — take out n(n+1) as common factor:
= n(n+1) · [⅓(2n+1) − ½]
= n(n+1) · [(4n+2 − 3)/6]

= n(n+1)(4n−1) / 6

Check

n=3: 1(1)+2(3)+3(5) = 1+6+15 = 22. Formula: 3·4·11/6 = 132/6 = 22. ✓

Partial Sums — Lower Limit Not 1

E

Sum from r = 4 to n

Find Σ r² for r = 4 to n, where n ≥ 4.

Method

Σ(r=4 to n) r² = Σ(r=1 to n) r² − Σ(r=1 to 3) r²

Calculate

Σ(r=1 to 3) r² = 1 + 4 + 9 = 14

Result

= ⅙n(n+1)(2n+1) − 14

Method of Differences

This is the most important technique in the P2 series topic. It works whenever you can write:

Key identity f(r) = g(r) − g(r−1) [or equivalently g(r+1) − g(r)] The sum then telescopes: all interior terms cancel, leaving only the first and last.

Why It Works — The Telescoping Argument

Write out the sum term by term. With f(r) = g(r) − g(r−1):

Telescoping expansion — terms cancel in pairs

r = 1: g(1) − g(0)

r = 2: g(2)* − g(1) (g(1) cancels with row above)

r = 3: g(3)* − g(2)

⋮ ⋮

r = n−1: g(n−1)* − g(n−2)

r = n: g(n) − g(n−1)

After cancellation: S = g(n) − g(0)

📌 Exam Technique — What Survives

If f(r) = g(r) − g(r−1), then Σ(r=1 to n) f(r) = g(n) − g(0)
If f(r) = g(r+1) − g(r), then Σ(r=1 to n) f(r) = g(n+1) − g(1)
For a partial sum Σ(r=m to n), the result is g(n) − g(m−1)
Always substitute a few values to confirm which terms survive before writing the final answer.

Finding the Difference Form — Partial Fractions Route

When f(r) is a rational function, use partial fractions to write it as a telescoping difference. The standard pattern is:

Typical rational form 1 / [r(r+1)] = 1/r − 1/(r+1) This telescopes perfectly. Sum from r=1 to n gives 1 − 1/(n+1) = n/(n+1).

Wider spacing 1 / [r(r+2)] = ½·[1/r − 1/(r+2)] Two terms survive at each end — be careful to count them correctly.

1

Decompose into partial fractions: write 1/[r(r+1)] = A/r + B/(r+1), solve for A and B.

2

Identify the telescoping structure: confirm it matches g(r) − g(r−1) or g(r+1) − g(r).

3

Write out the first 2–3 terms and last 2–3 terms to identify what survives. Never skip this step under exam conditions.

4

Simplify the surviving terms into a single fraction. Check with n=1 or n=2.

Sum to Infinity

If the series converges (each term → 0 as r → ∞), take the limit of your finite sum:

Example Σ(r=1 to ∞) 1/[r(r+1)] = lim[n→∞] n/(n+1) = 1

⛔ Do Not Assume Convergence

Only quote a sum to infinity if you can confirm the terms tend to zero. For example, Σ 1/r (harmonic series) diverges even though each term → 0. Always check the form of your closed-form sum as n → ∞.

Worked Examples

1

Method of Differences — Simple Rational

(i) Express 2/[r(r+2)] in partial fractions.
(ii) Hence find Σ 2/[r(r+2)] for r = 1 to n, and deduce the sum to infinity.

Part (i)

Let 2/[r(r+2)] = A/r + B/(r+2).
Multiply through: 2 = A(r+2) + Br
r = 0: 2 = 2A → A = 1
r = −2: 2 = −2B → B = −1

2/[r(r+2)] = 1/r − 1/(r+2)

Part (ii) Setup

S = Σ[1/r − 1/(r+2)] for r = 1 to n
Write out first three and last three terms:

Telescoping — spacing 2 means 2 terms survive at each end

r=1: 1/1−1/3

r=2: 1/2−1/4

r=3: 1/3−1/5

⋮ ⋮

r=n−1: 1/(n−1)−1/(n+1)

r=n: 1/n−1/(n+2)

Surviving terms: 1 + 1/2 − 1/(n+1) − 1/(n+2)

Simplify

S = 3/2 − 1/(n+1) − 1/(n+2)
= 3/2 − (2n+3)/[(n+1)(n+2)]

S = [3(n+1)(n+2) − 2(2n+3)] / [2(n+1)(n+2)]

Sum to ∞

As n → ∞, both 1/(n+1) → 0 and 1/(n+2) → 0.

S∞ = 3/2

2

Polynomial Difference — Non-Rational Form

Show that r(r+1)(r+2) − (r−1)r(r+1) = 3r(r+1).
Hence find Σ r(r+1) for r = 1 to n.

Show

LHS = r(r+1)[(r+2) − (r−1)] = r(r+1) · 3 = 3r(r+1). ✓

Identify

So r(r+1) = ⅓[r(r+1)(r+2) − (r−1)r(r+1)]
Let g(r) = r(r+1)(r+2). Then r(r+1) = ⅓[g(r) − g(r−1)].

Telescope

Σ r(r+1) = ⅓ · Σ[g(r) − g(r−1)] = ⅓[g(n) − g(0)]
g(n) = n(n+1)(n+2), g(0) = 0.

Σ r(r+1) = ⅓n(n+1)(n+2)

Verify

n=3: 1·2 + 2·3 + 3·4 = 2+6+12 = 20. Formula: ⅓·3·4·5 = 20. ✓

3

Cambridge-Style — Sum from r = 3 to 2n

Given that Σ(r=1 to n) r² = ⅙n(n+1)(2n+1), find Σ(r=3 to 2n) r² in terms of n, simplifying your answer.

Setup

Σ(r=3 to 2n) r² = Σ(r=1 to 2n) r² − Σ(r=1 to 2) r²

First sum

Replace n with 2n in the formula:
Σ(r=1 to 2n) r² = ⅙·(2n)(2n+1)(4n+1) = ⅓n(2n+1)(4n+1)

Subtract

Σ(r=1 to 2) r² = 1 + 4 = 5

Answer = ⅓n(2n+1)(4n+1) − 5

Practice Questions

All questions are at Cambridge 9231 examination level. Try each before revealing the hint or solution.

Question 1 — Standard Results

[4 marks]

Find Σ r(3r + 2) for r = 1 to n, simplifying your answer fully.

Expand r(3r+2) = 3r² + 2r, then apply Σr² and Σr formulae separately. Factorise the result by taking out n(n+1).

✓ Solution

Σ(3r² + 2r) = 3·⅙n(n+1)(2n+1) + 2·½n(n+1)
= ½n(n+1)(2n+1) + n(n+1)
= n(n+1)[½(2n+1) + 1]
= n(n+1)·½(2n+3)

= ½n(n+1)(2n+3)

Check n=2: 1·5 + 2·8 = 5+16 = 21. Formula: ½·2·3·7 = 21. ✓

Question 2 — Partial Fractions + Method of Differences

[6 marks]

(i) Express 4/[(2r−1)(2r+3)] in partial fractions.
(ii) Hence find Σ 4/[(2r−1)(2r+3)] for r = 1 to n.

For (i): write A/(2r−1) + B/(2r+3). Multiply through and substitute r = ½ and r = −3/2. For (ii): the gap between denominators is 4, so 2 terms survive at each end.

✓ Solution

Part (i):
4/[(2r−1)(2r+3)] = A/(2r−1) + B/(2r+3)
4 = A(2r+3) + B(2r−1)
r = ½: 4 = 4A → A = 1
r = −3/2: 4 = −4B → B = −1

= 1/(2r−1) − 1/(2r+3)

Part (ii):
Surviving terms (gap 4 = step of 2 in odd numbers, so 2 survive at each end):
r=1 gives 1/1; r=2 gives 1/3 — these survive at the start.
At the end: −1/(2n+1) and −1/(2n+3) survive.
S = 1 + 1/3 − 1/(2n+1) − 1/(2n+3)

= 4/3 − 1/(2n+1) − 1/(2n+3)

Question 3 — Sum to Infinity

[3 marks]

Using your answer to Question 2, or otherwise, find Σ(r=1 to ∞) 4/[(2r−1)(2r+3)].

Take the limit as n → ∞ of your closed-form expression. What happens to 1/(2n+1) and 1/(2n+3)?

✓ Solution

As n → ∞, both 1/(2n+1) → 0 and 1/(2n+3) → 0.

S∞ = 4/3

Question 4 — Polynomial Difference (Cambridge 9231 style)

[7 marks]

(i) Show that (r+1)⁴ − r⁴ = 4r³ + 6r² + 4r + 1.
(ii) By summing from r = 1 to n, use this result to derive the standard formula for Σr³.

For (ii): apply Method of Differences to the left side — it telescopes to (n+1)⁴ − 1. Then express 4Σr³ in terms of Σr², Σr, and Σ1 which you already know.

✓ Solution

Part (i): Expand (r+1)⁴ = r⁴ + 4r³ + 6r² + 4r + 1. Subtract r⁴. ✓

Part (ii):
Sum both sides from r=1 to n:
LHS telescopes: (n+1)⁴ − 1⁴ = (n+1)⁴ − 1
RHS: 4Σr³ + 6Σr² + 4Σr + Σ1
= 4Σr³ + 6·⅙n(n+1)(2n+1) + 4·½n(n+1) + n
= 4Σr³ + n(n+1)(2n+1) + 2n(n+1) + n

Rearrange:
4Σr³ = (n+1)⁴ − 1 − n(n+1)(2n+1) − 2n(n+1) − n
After careful expansion and factorisation:
4Σr³ = n²(n+1)²

Σr³ = ¼n²(n+1)² = [½n(n+1)]²

Interactive Series Calculator

Select a series type, set the upper limit n, and explore the partial sums and their closed-form values side by side.

Partial Sum Explorer

Σ r Sum of integers

Σ r² Sum of squares

Σ r³ Sum of cubes

Σ 1/r(r+1) Telescoping

Σ r(r+1) Product pairs

Σ (2r−1) Odd numbers

Upper limit n n = 8

Or enter n

36 Σ r = ½n(n+1) = ½·8·9 = 36

36 Total Sum

8 Last Term

4.5 Mean Term

— Growth Type

Formula Sheet — P2 Series

Quick reference for use during revision. These results are not provided in the Cambridge formula booklet — you must know them.

Standard Results

Σ 1 (r=1 to n)n

Σ r½n(n+1)

Σ r²⅙n(n+1)(2n+1)

Σ r³[½n(n+1)]²

Useful Derived Results

Σ r(r+1)⅓n(n+1)(n+2)

Σ r(r+1)(r+2)¼n(n+1)(n+2)(n+3)

Σ (2r−1)n²

Σ (2r−1)²⅓n(2n−1)(2n+1)

Method of Differences

f(r) = g(r)−g(r−1)⟹ S = g(n)−g(0)

f(r) = g(r+1)−g(r)⟹ S = g(n+1)−g(1)

Sum from r=m to ng(n)−g(m−1)

Common Telescoping PF

1/r(r+1)1/r − 1/(r+1)

1/r(r+2)½[1/r − 1/(r+2)]

1/(2r−1)(2r+1)½[1/(2r−1) − 1/(2r+1)]

Sum to ∞lim(n→∞) S_n

📋 Cambridge Exam Strategy — Summation of Series

Read the question structure: A "show that" part (i) always sets up part (ii). Use the given result exactly — don't re-derive.
Standard results: Expand f(r) as a polynomial first, then apply Σr², Σr, Σ1 individually.
Method of Differences: Always write out at least 3 terms from each end to identify surviving terms — one error here loses all subsequent marks.
Partial fractions route: Solve for A and B by substituting roots of the denominators (cover-up method is fastest).
Partial sums (r = k to n): Always subtract Σ(r=1 to k−1). Never guess at a new formula.
Sum to infinity: State what happens to each remaining term as n → ∞ before quoting the limit.