Sigma Notation

Sigma notation (∑) provides a compact way to write the sum of a sequence of terms. The Greek capital letter sigma (∑) means "sum". This notation is used extensively in A-Level Mathematics (9MA0) and understanding it is essential for work with series.

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Key Notation

The expression ∑ from r = 1 to n of f(r) means: add up f(r) for r = 1, 2, 3, ..., n.

Component	Meaning
∑	"The sum of"
r	The index variable (also called the dummy variable)
r = 1 (below ∑)	The starting value of r
n (above ∑)	The final value of r
f(r)	The expression to be summed

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Writing Sums in Sigma Notation

Worked Example 1

Write the sum 2 + 4 + 6 + 8 + ... + 100 using sigma notation.

Each term is 2r, where r goes from 1 to 50.

∑ (r = 1 to 50) 2r

Worked Example 2

Write the sum 1 + 4 + 9 + 16 + ... + 400 using sigma notation.

Each term is r², where r goes from 1 to 20.

∑ (r = 1 to 20) r²

Worked Example 3

Write the sum 3 + 5 + 7 + 9 + ... + 41 in sigma notation.

The rth term is 2r + 1. When r = 1: 3. When r = 20: 41.

∑ (r = 1 to 20) (2r + 1)

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Evaluating Sums Given in Sigma Notation

Worked Example 4

Evaluate ∑ (r = 1 to 5) (3r - 1).

Write out each term:

• r = 1: 3(1) - 1 = 2
• r = 2: 3(2) - 1 = 5
• r = 3: 3(3) - 1 = 8
• r = 4: 3(4) - 1 = 11
• r = 5: 3(5) - 1 = 14

Sum = 2 + 5 + 8 + 11 + 14 = 40

Alternatively: this is an AP with a = 2, d = 3, n = 5. S(5) = 5/2 x (2 + 14) = 5/2 x 16 = 40. Same answer.

Worked Example 5

Evaluate ∑ (r = 1 to 4) r³.

= 1³ + 2³ + 3³ + 4³ = 1 + 8 + 27 + 64 = 100

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Standard Results

You should know these standard results (they are in the formula booklet):

Sum	Result
∑ (r = 1 to n) 1	n
∑ (r = 1 to n) r	n(n + 1)/2
∑ (r = 1 to n) r²	n(n + 1)(2n + 1)/6
∑ (r = 1 to n) r³	[n(n + 1)/2]²

Worked Example 6

Find ∑ (r = 1 to 100) r.

Using the formula: 100 x 101 / 2 = 5050

Worked Example 7

Find ∑ (r = 1 to 20) r².

Using the formula: 20 x 21 x 41 / 6 = 17220 / 6 = 2870

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Properties of Sigma Notation

These rules allow you to break down complex sums:

Rule 1: Constant factor — ∑ c x f(r) = c x ∑ f(r)

Rule 2: Sum/difference — ∑ [f(r) ± g(r)] = ∑ f(r) ± ∑ g(r)

Rule 3: Constant sum — ∑ (r = 1 to n) c = cn

Worked Example 8

Find ∑ (r = 1 to 50) (4r² + 3r - 2).

Split into parts: = 4 x ∑r² + 3 x ∑r - ∑2

= 4 x [50 x 51 x 101 / 6] + 3 x [50 x 51 / 2] - 2 x 50

= 4 x 42925 + 3 x 1275 - 100

= 171700 + 3825 - 100

= 175425

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Changing the Starting Index

Sometimes the sum does not start at r = 1. To handle this:

∑ (r = m to n) f(r) = ∑ (r = 1 to n) f(r) - ∑ (r = 1 to m-1) f(r)

Worked Example 9

Find ∑ (r = 11 to 30) r.

= ∑ (r = 1 to 30) r - ∑ (r = 1 to 10) r

= 30 x 31 / 2 - 10 x 11 / 2

= 465 - 55

= 410

Worked Example 10

Find ∑ (r = 5 to 20) (2r + 1).

= ∑ (r = 1 to 20) (2r + 1) - ∑ (r = 1 to 4) (2r + 1)

For ∑ (r = 1 to 20) (2r + 1): = 2 x 20 x 21 / 2 + 20 = 420 + 20 = 440

For ∑ (r = 1 to 4) (2r + 1): = 2 x 4 x 5 / 2 + 4 = 20 + 4 = 24

Answer: 440 - 24 = 416

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Exam Tips

Tip	Detail
Write out terms	If unsure, write out the first few terms to check your sigma expression
Use standard results	For sums involving r, r², r³, use the formulae rather than adding term by term
Starting index	If the sum starts at r = k (not 1), split into two sums starting from r = 1
Dummy variable	The letter used (r, k, i) does not matter
Factorise	After using standard results, look to simplify and factorise your final answer

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Practice Problems

1. Write 5 + 10 + 15 + ... + 200 in sigma notation. [Answer: ∑ (r = 1 to 40) 5r]
2. Evaluate ∑ (r = 1 to 6) (2r² - 1). [Answer: 176]
3. Find ∑ (r = 1 to 30) r². [Answer: 9455]
4. Find ∑ (r = 6 to 15) r. [Answer: 105]
5. Find ∑ (r = 1 to n) (6r - 1) in terms of n. [Answer: n(3n + 2)]

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Summary

• ∑ notation is shorthand for "add up all the terms".
• Standard results for ∑r, ∑r², ∑r³ are in the formula booklet.
• You can factor out constants and split sums.
• To handle a sum starting at r = m, subtract the sum from r = 1 to m - 1.

A-Level Deep Dive: Sigma Notation

Spec mapping

Edexcel 9MA0 specification, Pure Mathematics — Sequences and series covers sigma notation for sums of series (refer to the official specification document for exact wording). Sigma notation is the language in which every series result on 9MA0 is written, so although the standalone sub-strand is small, fluency here is load-bearing across the whole specification. The Edexcel formula booklet provides arithmetic and geometric series sums in closed form; the standard results $\sum_{r=1}^n r = \tfrac{1}{2}n(n+1)$∑r=1n​r=21​n(n+1) and $\sum_{r=1}^n r^2 = \tfrac{1}{6}n(n+1)(2n+1)$∑r=1n​r2=61​n(n+1)(2n+1) are not on the AS booklet but appear in the Year 2 / Further Mathematics booklet. AS candidates can derive the linear result from the arithmetic series formula; the quadratic result is taken as standard at A2 and Further. Sigma notation is examined synoptically with arithmetic and geometric series (where $S_n = \sum_{r=1}^n a_r$Sn​=∑r=1n​ar​), proof by induction (Further Mathematics — proving $\sum r^2$∑r2 etc.), binomial expansion ($(1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$(1+x)n=∑r=0n​(rn​)xr), integration as the continuous analogue ($\int$∫ ↔ $\Sigma$Σ as Riemann sum limit), and statistics ($\sum x_i$∑xi​, $\sum x_i^2$∑xi2​ underlie the mean and variance formulae printed in the formula booklet).

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Worked example with full mark scheme

Question (8 marks): Evaluate $\displaystyle\sum_{r=1}^{20}\bigl(3r^2 - 2r + 5\bigr)$r=1∑20​(3r2−2r+5), using the standard results $\sum_{r=1}^n r = \tfrac{1}{2}n(n+1)$∑r=1n​r=21​n(n+1) and $\sum_{r=1}^n r^2 = \tfrac{1}{6}n(n+1)(2n+1)$∑r=1n​r2=61​n(n+1)(2n+1).

Solution with mark scheme:

Step 1 — split the sum using linearity.

$\sum_{r=1}^{20}\bigl(3r^2 - 2r + 5\bigr) = 3\sum_{r=1}^{20} r^2 - 2\sum_{r=1}^{20} r + \sum_{r=1}^{20} 5$∑r=120​(3r2−2r+5)=3∑r=120​r2−2∑r=120​r+∑r=120​5

M1 — splitting the sum into three pieces and pulling out the constant multiples $3$3 and $-2$−2. The justification is that $\Sigma$Σ is a linear operator: $\sum (af(r) + bg(r)) = a\sum f(r) + b\sum g(r)$∑(af(r)+bg(r))=a∑f(r)+b∑g(r). Common error: students try to "factor" the polynomial under the sigma, e.g. writing $\sum (3r^2 - 2r + 5) = (3r-?)(r-?)$∑(3r2−2r+5)=(3r−?)(r−?), treating the sum like an algebraic expression. Loses M1 instantly.

A1 — correctly written split with constants out front.

Step 2 — apply the standard result for $\sum r^2$∑r2 with $n = 20$n=20.

$\sum_{r=1}^{20} r^2 = \tfrac{1}{6}(20)(21)(41) = \tfrac{1}{6} \cdot 17{,}220 = 2870$∑r=120​r2=61​(20)(21)(41)=61​⋅17,220=2870

M1 — substitution into $\tfrac{1}{6}n(n+1)(2n+1)$61​n(n+1)(2n+1) with $n=20$n=20, giving $2n+1 = 41$2n+1=41.

A1 — correct value $2870$2870.

Step 3 — apply the standard result for $\sum r$∑r with $n = 20$n=20.

$\sum_{r=1}^{20} r = \tfrac{1}{2}(20)(21) = 210$∑r=120​r=21​(20)(21)=210

M1 — substitution into $\tfrac{1}{2}n(n+1)$21​n(n+1).

Step 4 — handle the constant sum.

$\sum_{r=1}^{20} 5 = 5 \cdot 20 = 100$∑r=120​5=5⋅20=100

M1 — recognising that $\sum_{r=1}^{n} c = cn$∑r=1n​c=cn for a constant $c$c. Common error: writing $\sum_{r=1}^{20} 5 = 5$∑r=120​5=5 (forgetting the multiplicity) or $5 \cdot 21$5⋅21 (off-by-one).

Step 5 — combine.

$3(2870) - 2(210) + 100 = 8610 - 420 + 100 = 8290$3(2870)−2(210)+100=8610−420+100=8290

A1 — correct final value $8290$8290.

A1 (presentation) — the answer is given as a single integer with full working visible; no decimal approximation, no unsimplified intermediate fractions left behind.

Total: 8 marks (M4 A4).

Examiner notes on the worked solution: A candidate who writes the first line — splitting $\sum (3r^2 - 2r + 5) = 3\sum r^2 - 2\sum r + \sum 5$∑(3r2−2r+5)=3∑r2−2∑r+∑5 — has effectively secured the first two marks regardless of arithmetic slips later. This is why marking the structure first matters more than computing fast. The arithmetic at $\tfrac{1}{6}(20)(21)(41)$61​(20)(21)(41) rewards a candidate who multiplies in the order $(20)(21) = 420$(20)(21)=420, then $420 \times 41 = 17{,}220$420×41=17,220, then divides by $6$6 — rather than the temptation to compute $21 \times 41$21×41 first, which is $861$861 and harder to multiply by $20$20 accurately under timed pressure. Edexcel mark schemes typically award method marks generously: substituting the correct $n$n into a clearly identified standard result earns M1 even if the subsequent arithmetic is wrong. Accuracy marks (A1) are reserved for the final correct numerical answer. So a candidate who mis-computes $420 \times 41 = 17{,}220$420×41=17,220 as $17{,}210$17,210 would still earn M3 but lose A2 — a distinction that matters: the question is worth 8 marks but a careful candidate with the wrong final answer might still score 6/8.

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Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): Show that $\displaystyle\sum_{r=1}^{n}(2r-1)^2 = \tfrac{1}{3}n(2n-1)(2n+1)$r=1∑n​(2r−1)2=31​n(2n−1)(2n+1) for all positive integers $n$n.

Mark scheme decomposition by AO:

• M1 (AO1.1a) — expanding $(2r-1)^2 = 4r^2 - 4r + 1$(2r−1)2=4r2−4r+1 under the sigma.
• M1 (AO1.1b) — splitting $\sum (4r^2 - 4r + 1) = 4\sum r^2 - 4\sum r + \sum 1$∑(4r2−4r+1)=4∑r2−4∑r+∑1.
• M1 (AO1.1b) — substituting $\sum r^2 = \tfrac{1}{6}n(n+1)(2n+1)$∑r2=61​n(n+1)(2n+1), $\sum r = \tfrac{1}{2}n(n+1)$∑r=21​n(n+1), $\sum 1 = n$∑1=n.
• A1 (AO1.1b) — combining: $\tfrac{2}{3}n(n+1)(2n+1) - 2n(n+1) + n$32​n(n+1)(2n+1)−2n(n+1)+n.
• M1 (AO2.1) — factorising out $\tfrac{1}{3}n$31​n and simplifying the bracket: $\tfrac{1}{3}n\bigl[2(n+1)(2n+1) - 6(n+1) + 3\bigr]$31​n[2(n+1)(2n+1)−6(n+1)+3].
• A1 (AO2.1) — expanding the inner bracket and recognising it as $(2n-1)(2n+1)$(2n−1)(2n+1), giving the printed result.

Total: 6 marks split AO1 = 4, AO2 = 2. The AO2 marks are reserved for the algebraic manipulation that converts the raw substituted form into the factorised target; an unfactorised numerically-correct expression earns the M marks but loses both A2 marks. A common candidate trap on this style of question is to verify the formula numerically (e.g. checking $n=3$n=3 gives $1+9+25 = 35 = \tfrac{1}{3}(3)(5)(7)$1+9+25=35=31​(3)(5)(7)) and stop there — but "show that" is a general proof requirement, not a verification, and substituting a single value scores zero. The expected route is the algebraic chain shown above, with explicit $\Sigma$Σ manipulation visible at every stage. A cleaner alternative for A* candidates: induction. Base case $n=1$n=1: LHS $= 1$=1, RHS $= \tfrac{1}{3}(1)(1)(3) = 1$=31​(1)(1)(3)=1. Inductive step: assume $\sum_{r=1}^{k}(2r-1)^2 = \tfrac{1}{3}k(2k-1)(2k+1)$∑r=1k​(2r−1)2=31​k(2k−1)(2k+1) and add $(2(k+1)-1)^2 = (2k+1)^2$(2(k+1)−1)2=(2k+1)2 to both sides, then factorise. Either method earns full marks; the algebraic route is more typical of 9MA0, the inductive route of Further Mathematics 9FM0.

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Synoptic links

Connects to:

1. Arithmetic and geometric series: $S_n = \sum_{r=1}^n a_r$Sn​=∑r=1n​ar​ is the definition of a partial sum. The closed forms $S_n = \tfrac{n}{2}(2a + (n-1)d)$Sn​=2n​(2a+(n−1)d) (arithmetic) and $S_n = a(1-r^n)/(1-r)$Sn​=a(1−rn)/(1−r) (geometric) are evaluations of these sigma expressions — once you write the series in sigma form, you choose which closed form applies.

2. Proof by induction (Further Mathematics): the standard results $\sum_{r=1}^n r^k$∑r=1n​rk for $k = 1, 2, 3$k=1,2,3 are typically proved by induction. The case $k=3$k=3 (Faulhaber's formula $\sum r^3 = (\tfrac{1}{2}n(n+1))^2 = (\sum r)^2$∑r3=(21​n(n+1))2=(∑r)2) is a favourite Year 2 question.

3. Integration as continuous summation: $\int_a^b f(x)\,dx = \lim_{n \to \infty} \sum_{r=1}^n f(x_r)\,\Delta x$∫ab​f(x)dx=limn→∞​∑r=1n​f(xr​)Δx where $\Delta x = (b-a)/n$Δx=(b−a)/n. The integral is the limit of a Riemann sum — sigma notation is integration's discrete twin.

4. Statistics — formula booklet: the mean $\bar{x} = \tfrac{1}{n}\sum x_i$xˉ=n1​∑xi​ and variance $\sigma^2 = \tfrac{1}{n}\sum x_i^2 - \bar{x}^2$σ2=n1​∑xi2​−xˉ2 are sigma expressions. The summary statistics $S_{xx} = \sum x_i^2 - \tfrac{(\sum x_i)^2}{n}$Sxx​=∑xi2​−n(∑xi​)2​ used in regression and correlation are pure sigma manipulation.

5. Combinatorics and the binomial theorem: $(1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r$(1+x)n=∑r=0n​(rn​)xr. Substituting $x=1$x=1 gives $\sum \binom{n}{r} = 2^n$∑(rn​)=2n; substituting $x=-1$x=−1 gives $\sum (-1)^r \binom{n}{r} = 0$∑(−1)r(rn​)=0. Both are sigma-notation identities derived by clever choice of $x$x.

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Mark-scheme literacy

Sigma-notation questions on 9MA0 split AO marks as:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Splitting sums correctly, substituting standard results with the correct $n$n, evaluating arithmetic
AO2 (reasoning / interpretation)	25–35%	Manipulating limits (e.g. converting $\sum_{r=k}^n$∑r=kn​ to $\sum_{r=1}^n - \sum_{r=1}^{k-1}$∑r=1n​−∑r=1k−1​), factorising into the printed target form, justifying linearity
AO3 (problem-solving)	0–10%	Open-modelling sigma problems are rare at AS; appear in Year 2 with telescoping or induction contexts