Sigma Notation

Understanding and using summation notation for series

SPEC 4.3 • SEQUENCES AND SERIES

What is Sigma Notation?

Definition

Sigma notation (also called summation notation) is a compact way to write the sum of a series of terms. The Greek capital letter sigma (Σ) means "sum".

r = 1 \sum n u_{r} = u_{1} + u_{2} + u_{3} + \dots + u_{n}

Instead of writing out every term in a long sum, we use sigma notation to express the sum concisely.

Example

r = 1 \sum 5 r = 1 + 2 + 3 + 4 + 5 = 15

This means "sum all values of r from r = 1 to r = 5"

Components of Sigma Notation

r = a \sum b f (r)

ComponentNameMeaning

Σ

Sigma (sum symbol)Indicates we are adding terms

r = a

Lower limitStarting value of the index (can be any integer)

b

Upper limitEnding value of the index

r

Index variableThe variable being summed over

f (r)

General termExpression to evaluate for each r

Important

The lower limit doesn't have to be 1. It can start from any integer including 0, 2, 3, or even negative numbers!

Evaluating Sigma Notation

Worked Example01

Evaluate

r = 1 \sum 4 (2 r + 1)

Step 1 Write out each term by substituting r = 1, 2, 3, 4

When r = 1: 2(1) + 1 = 3

When r = 2: 2(2) + 1 = 5

When r = 3: 2(3) + 1 = 7

When r = 4: 2(4) + 1 = 9

Step 2 Add all the terms

r = 1 \sum 4 (2 r + 1) = 3 + 5 + 7 + 9 = 24

Method: Evaluating Sigma Sums

Systematic Method

Identify the limits: Note the starting value (lower limit) and ending value (upper limit)

Substitute each value: Replace the index variable with each integer from lower to upper limit

Evaluate each term: Calculate the value of the expression for each substitution

Sum all terms: Add all the calculated values together

Exam Tip

Always write out at least the first few and last terms to check you understand the pattern. This helps avoid mistakes with the limits.

Lower Limit Starting from 0

Worked Example02

Evaluate

r = 0 \sum 4 (r^{2} + 1)

Step 1 The sum starts at r = 0 (not r = 1)

When r = 0: 0² + 1 = 1

When r = 1: 1² + 1 = 2

When r = 2: 2² + 1 = 5

When r = 3: 3² + 1 = 10

When r = 4: 4² + 1 = 17

Step 2 Add all the terms (5 terms total)

r = 0 \sum 4 (r^{2} + 1) = 1 + 2 + 5 + 10 + 17 = 35

Different Starting Limits

Worked Example03

Evaluate

r = 3 \sum 7 (2 r - 1)

Step 1 The sum starts at r = 3

When r = 3: 2(3) - 1 = 5

When r = 4: 2(4) - 1 = 7

When r = 5: 2(5) - 1 = 9

When r = 6: 2(6) - 1 = 11

When r = 7: 2(7) - 1 = 13

Step 2 Add all the terms

r = 3 \sum 7 (2 r - 1) = 5 + 7 + 9 + 11 + 13 = 45

Writing Series in Sigma Notation

To convert a series into sigma notation, we need to:

Method

Identify the pattern: Find a formula for the general term in terms of r

Determine the limits: Find the starting and ending values of r

Write in sigma form: Combine using Σ notation

Common Patterns

1 + 2 + 3 + \dots + n 1^{2} + 2^{2} + 3^{2} + \dots + n^{2} 2 + 4 + 6 + \dots + 2 n \to r = 1 \sum n r \to r = 1 \sum n r^{2} \to r = 1 \sum n 2 r

Converting to Sigma Notation

Worked Example04

Write the series 3 + 5 + 7 + 9 + 11 + 13 in sigma notation.

Step 1 Identify the pattern

This is an arithmetic sequence with first term 3 and common difference 2.

The rth term: when r = 1, term = 3 = 2(1) + 1

General term:

2 r + 1

Step 2 Find the limits

First term (r = 1): 2(1) + 1 = 3 ✓

Last term: 13 = 2r + 1, so r = 6

Answer

r = 1 \sum 6 (2 r + 1)

Sum of First n Natural Numbers

Key Formula (In Formula Booklet)

r = 1 \sum n r = 1 + 2 + 3 + \dots + n = \frac{n ( n + 1 )}{2}

This formula is given in the exam formula booklet and is essential for many sigma notation questions.

Example 1

r = 1 \sum 10 r = \frac{10 \times 11}{2} = 55

Example 2

r = 1 \sum 100 r = \frac{100 \times 101}{2} = 5050

Proof Hint

You should know the proof: write the sum forwards and backwards, add them together to get n copies of (n + 1), then divide by 2.

Using the Natural Numbers Formula

Worked Example05

Find the value of

r = 1 \sum 50 r

Step 1 Apply the formula

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

Here n = 50

Step 2 Substitute and calculate

r = 1 \sum 50 r = \frac{50 \times 51}{2} = \frac{2550}{2} = 1275

Properties of Sigma Notation

Property 1: Constant Multiple

r = 1 \sum n c f (r) = c r = 1 \sum n f (r)

A constant can be factored out of a sum.

Property 2: Sum of Sums

r = 1 \sum n [f (r) + g (r)] = r = 1 \sum n f (r) + r = 1 \sum n g (r)

A sum can be split into separate sums.

Property 3: Sum of a Constant

r = 1 \sum n c = n c

Adding a constant n times gives n × c.

Using Properties

Worked Example06

Find the value of

r = 1 \sum 20 (3 r + 2)

Step 1 Split the sum using properties

r = 1 \sum 20 (3 r + 2) = r = 1 \sum 20 3 r + r = 1 \sum 20 2

Step 2 Factor out constants

= 3 r = 1 \sum 20 r + 2 \times 20

Step 3 Use the formula for sum of natural numbers

= 3 \times \frac{20 \times 21}{2} + 40 = 3 \times 210 + 40 = 630 + 40 = 670

Splitting and Combining Sums

Splitting a Sum

r = 1 \sum n f (r) = r = 1 \sum k f (r) + r = k + 1 \sum n f (r)

This property is useful when we need to find a sum that doesn't start from r = 1.

Key Technique

To find

r = m \sum n f (r)

where m > 1:

r = m \sum n f (r) = r = 1 \sum n f (r) - r = 1 \sum m - 1 f (r)

Example

r = 5 \sum 10 r = r = 1 \sum 10 r - r = 1 \sum 4 r = 55 - 10 = 45

Sums Not Starting from 1

Worked Example07

Find the value of

r = 11 \sum 30 r

Step 1 Use the splitting technique

r = 11 \sum 30 r = r = 1 \sum 30 r - r = 1 \sum 10 r

Step 2 Calculate each sum using the formula

r = 1 \sum 30 r = \frac{30 \times 31}{2} = 465

r = 1 \sum 10 r = \frac{10 \times 11}{2} = 55

Step 3 Subtract

r = 11 \sum 30 r = 465 - 55 = 410

Algebraic Expression in n

Worked Example08

Given that

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

, find

r = 1 \sum n (4 r - 3)

in terms of n.

Step 1 Split the sum

r = 1 \sum n (4 r - 3) = r = 1 \sum n 4 r - r = 1 \sum n 3 = 4 r = 1 \sum n r - 3 n

Step 2 Substitute the formula

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

Step 3 Simplify

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

Practice

1 Evaluate

r = 0 \sum 5 (2 r + 3)

Step 1 Write out each term (starting from r = 0)

r = 0: 2(0) + 3 = 3

r = 1: 2(1) + 3 = 5

r = 2: 2(2) + 3 = 7

r = 3: 2(3) + 3 = 9

r = 4: 2(4) + 3 = 11

r = 5: 2(5) + 3 = 13

Step 2 Sum = 3 + 5 + 7 + 9 + 11 + 13 = 48

Practice

2 Write the series 4 + 7 + 10 + 13 + 16 + 19 in sigma notation.

Step 1 Identify the pattern: first term = 4, common difference = 3

General term: 3r + 1 (check: when r = 1, 3(1) + 1 = 4 ✓)

Step 2 Find upper limit: 19 = 3r + 1, so r = 6

Answer

r = 1 \sum 6 (3 r + 1)

Practice

3 Find the value of

r = 6 \sum 25 r

Step 1 Use the splitting technique (sum doesn't start from 1)

r = 6 \sum 25 r = r = 1 \sum 25 r - r = 1 \sum 5 r

Step 2 Calculate each sum

r = 1 \sum 25 r = \frac{25 \times 26}{2} = 325

r = 1 \sum 5 r = \frac{5 \times 6}{2} = 15

Answer 325 - 15 = 310

Summary

Sigma Notation

ConceptFormula/Rule

Basic form

r = a \sum b f (r) = f (a) + f (a + 1) + \dots + f (b)

Sum of natural numbers

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

Constant multiple

\sum c f (r) = c \sum f (r)

Sum of sums

\sum [f (r) + g (r)] = \sum f (r) + \sum g (r)

Sum of constant

r = 1 \sum n c = n c

Splitting sums

r = m \sum n = r = 1 \sum n - r = 1 \sum m - 1

Key Exam Skills

Evaluate sigma sums with any lower limit (including r = 0)

Write series in sigma notation

Use the formula for sum of first n natural numbers

Apply properties to simplify complex sums

Use the splitting technique for sums not starting from r = 1