1.4.7 Sigma notation (A-level only) - Revision Guide

What you'll learn

How sigma notation represents the sum of a sequence of terms.
How to expand and evaluate a sum written using sigma notation.
How to write a given series compactly using sigma notation.
How to manipulate sums and use standard summation formulae.

From sequences to series

A sequence is an ordered list of numbers. Each number in the sequence is called a term.

For example, the sequence

9,…3,\ 5,\ 7,\ 9,\ldots

has nth term $2 n + 1$ , where $n$ is the position of a term. Substituting $n = 1$ gives the first term, substituting $n = 2$ gives the second term, and so on.

A series is formed by adding the terms of a sequence:

3+5+7+9+⋯3+5+7+9+\cdots

Sigma notation provides a concise way to write such additions.

Understanding sigma notation

The Greek capital letter sigma, $∑\sum$ , means add up or find the sum of.

A typical sum looks like this:

∑r=15(2r+1)\sum_{r=1}^{5}(2r+1)

Definition

Parts of sigma notation

In the expression $∑r=15(2r+1)\displaystyle \sum_{r=1}^{5}(2r+1)$ :

$∑\sum$ is the instruction to add.
$r$ is the index variable, which keeps track of the terms.
$r = 1$ is the lower limit, so the first value used is 1.
The number above the sigma is the upper limit, so the final value used is 5.
$2 r + 1$ is the summand, meaning the expression whose values are added.

To expand the sum, substitute each whole-number value of $r$ from the lower limit to the upper limit, including both limits.

Example

Expanding a sigma sum

Expand $∑r=15(2r+1)\displaystyle \sum_{r=1}^{5}(2r+1)$ .

The index runs through $r = 1, 2, 3, 4, 5$ , so substitute each of these values into $2 r + 1$ .
This gives $∑r=15(2r+1)=(2(1)+1)+(2(2)+1)+(2(3)+1)+(2(4)+1)+(2(5)+1).\begin{aligned} \sum_{r=1}^{5}(2r+1) &=(2(1)+1)+(2(2)+1)+(2(3)+1)\\ &\quad +(2(4)+1)+(2(5)+1). \end{aligned}$
Simplifying each term gives $3 + 5 + 7 + 9 + 11 = 35.$

Common Mistake

Leaving out an endpoint

Both limits are included. If the index runs from 1 to 5, there are five terms, not four.

Counting the number of terms

If the index runs through every integer from $a$ to $b$ , the number of terms is

b - a + 1.

For example, $∑r=410r\displaystyle \sum_{r=4}^{10}r$ contains $10 - 4 + 1 = 7$ terms.

Key Idea

Limits are values, not term counts

The upper limit gives the final value of the index. It does not necessarily give the number of terms unless the lower limit is 1.

Evaluating sums directly

For a short sum, the safest method is often to expand it and calculate each term.

Be especially careful when the summand contains a power or brackets. For example, in $r+1)^2$ , you must square the whole value of $r + 1$ .

Example

Evaluating a sum with squares

Evaluate $∑r=25(r+1)2\displaystyle \sum_{r=2}^{5}(r+1)^2$ .

The index values are $r = 2, 3, 4, 5$ , giving $2+1)^2+(3+1)^2+(4+1)^2+(5+1)^2.$
Evaluate each squared term: $3^2+4^2+5^2+6^2=9+16+25+36.$
Add the values to obtain $∑r=25(r+1)2=86.\sum_{r=2}^{5}(r+1)^2=86.$

Tip

Check the first and last terms

Before adding, substitute the lower limit and upper limit separately. This quickly checks that your expansion starts and finishes with the correct terms.

Writing a series using sigma notation

To convert an expanded series into sigma notation, you need to identify:

a formula that generates each term;
a suitable index variable;
the starting and finishing values of that index.

The index letter is a placeholder. Letters such as $r$ , $k$ and $i$ are commonly used, and changing the letter does not change the sum:

∑r=1nr2=∑k=1nk2.\sum_{r=1}^{n}r^2=\sum_{k=1}^{n}k^2.

Example

Writing an arithmetic series compactly

Write $5+8+11+⋯+355+8+11+\cdots+35$ using sigma notation.

The terms increase by 3, so they form an arithmetic sequence. Starting with index $r = 1$ , its rth term is $5 + 3 (r - 1) = 3 r + 2.$
Find the index of the final term by solving $3 r + 2 = 35,$ which gives $r = 11$ .
Therefore the series is $∑r=111(3r+2).\sum_{r=1}^{11}(3r+2).$

There may be more than one correct way to represent the same series. For example,

5+8+11+⋯+355+8+11+\cdots+35

can also be written as

∑r=010(3r+5).\sum_{r=0}^{10}(3r+5).

The limits and the summand have both changed, but the generated terms are identical.

Common Mistake

Changing limits without changing the formula

If you shift the starting value of the index, you usually need to adjust the summand as well. Always expand the first two terms and the last term to check your notation.

Algebra with sigma notation

Sigma notation follows the usual rules of addition and multiplication. Sums with the same limits can be combined:

∑r=1nar+∑r=1nbr=∑r=1n(ar+br).\sum_{r=1}^{n}a_r+\sum_{r=1}^{n}b_r = \sum_{r=1}^{n}(a_r+b_r).

A constant factor can be taken outside a sum:

∑r=1ncar=c∑r=1nar.\sum_{r=1}^{n}ca_r = c\sum_{r=1}^{n}a_r.

However, adding a constant inside a sum adds that constant once for every value of the index:

∑r=1nc=cn.\sum_{r=1}^{n}c=cn.

Consequently,

∑r=1n(ar+b)=a∑r=1nr+bn.\sum_{r=1}^{n}(ar+b) = a\sum_{r=1}^{n}r+bn.

Example

Splitting and evaluating a sum

Evaluate $∑r=120(3r−2)\displaystyle \sum_{r=1}^{20}(3r-2)$ .

Split the summand and take the constant factor outside: $∑r=120(3r−2)=3∑r=120r−∑r=1202.\sum_{r=1}^{20}(3r-2) = 3\sum_{r=1}^{20}r-\sum_{r=1}^{20}2.$
Use $∑r=120r=20(21)2=210\displaystyle \sum_{r=1}^{20}r=\frac{20(21)}{2}=210$ , and note that the constant 2 is added 20 times: $∑r=1202=2(20)=40.\sum_{r=1}^{20}2=2(20)=40.$
Therefore, $3 (210) - 40 = 590.$

Common Mistake

Treating a constant as a single term

In $∑r=1n(r+4)\displaystyle \sum_{r=1}^{n}(r+4)$ , the 4 occurs in every term. Its total contribution is $4 n$ , not 4.

Standard summation formulae

For large upper limits, expanding every term is inefficient. The following standard results are useful:

∑r=1n1=n,∑r=1nr=n(n+1)2,∑r=1nr2=n(n+1)(2n+1)6,∑r=1nr3=(n(n+1)2)2.\begin{aligned} \sum_{r=1}^{n}1&=n,\\ \sum_{r=1}^{n}r&=\frac{n(n+1)}{2},\\ \sum_{r=1}^{n}r^2&=\frac{n(n+1)(2n+1)}{6},\\ \sum_{r=1}^{n}r^3&=\left(\frac{n(n+1)}{2}\right)^2. \end{aligned}

These formulae apply directly when the lower limit is 1. If the lower limit is different, subtract the unwanted beginning of the sum.

For example,

∑r=4nr2=∑r=1nr2−∑r=13r2.\sum_{r=4}^{n}r^2 = \sum_{r=1}^{n}r^2-\sum_{r=1}^{3}r^2.

Example

Using a standard sum with a shifted lower limit

Evaluate $∑r=512r2\displaystyle \sum_{r=5}^{12}r^2$ .

Rewrite the required sum as the sum from 1 to 12 minus the sum from 1 to 4: $∑r=512r2=∑r=112r2−∑r=14r2.\sum_{r=5}^{12}r^2 = \sum_{r=1}^{12}r^2-\sum_{r=1}^{4}r^2.$
Apply the square-sum formula: $12(13)(25)6−4(5)(9)6.\frac{12(13)(25)}{6}-\frac{4(5)(9)}{6}.$
Calculate the difference: $650 - 30 = 620.$

Common Mistake

Check the required lower limit

To find a sum beginning at $r = 5$ , subtract the terms up to and including $r = 4$ . Subtracting up to $r = 5$ would remove one term too many.

Sigma notation and known series

Sigma notation can also describe arithmetic and geometric series. For example,

∑r=1n(a+(r−1)d)\sum_{r=1}^{n}\bigl(a+(r-1)d\bigr)

represents the first $n$ terms of an arithmetic sequence with first term $a$ and common difference $d$ .

Similarly,

r−1\sum_{r=1}^{n}ar^{\,r-1}

would be ambiguous because the same letter is being used as both the index and a fixed common ratio. It is clearer to use a different index:

∑k=1naqk−1,\sum_{k=1}^{n}aq^{k-1},

where $a$ is the first term and $q$ is the common ratio.

Key Idea

The index is a changing variable

Quantities such as $a$ , $d$ , $q$ and $n$ remain fixed while the index runs through its stated integer values.

Exam technique

In the exam

Read the lower limit, upper limit and summand separately before doing any calculation.
For an unfamiliar sum, write out its first two terms and final term to expose indexing errors.
Use standard summation formulae for large sums, but adjust carefully when the lower limit is not 1.
When constructing sigma notation, expand your answer briefly to check that it reproduces the original series.

Self review

Check yourself

Can you expand and evaluate $∑r=26(3r−1)\displaystyle \sum_{r=2}^{6}(3r-1)$ ?
How would you write $4+9+14+⋯+494+9+14+\cdots+49$ using sigma notation?
How could you evaluate $∑r=720r2\displaystyle \sum_{r=7}^{20}r^2$ without listing all fourteen terms?

What you'll learn

How sigma notation represents the sum of a sequence of terms.
How to expand and evaluate a sum written using sigma notation.
How to write a given series compactly using sigma notation.
How to manipulate sums and use standard summation formulae.

From sequences to series

A sequence is an ordered list of numbers. Each number in the sequence is called a term.

For example, the sequence

9,…3,\ 5,\ 7,\ 9,\ldots

has nth term $2 n + 1$ , where $n$ is the position of a term. Substituting $n = 1$ gives the first term, substituting $n = 2$ gives the second term, and so on.

A series is formed by adding the terms of a sequence:

3+5+7+9+⋯3+5+7+9+\cdots

Sigma notation provides a concise way to write such additions.

Understanding sigma notation

The Greek capital letter sigma, $∑\sum$ , means add up or find the sum of.

A typical sum looks like this:

∑r=15(2r+1)\sum_{r=1}^{5}(2r+1)

Definition

Parts of sigma notation

In the expression $∑r=15(2r+1)\displaystyle \sum_{r=1}^{5}(2r+1)$ :

$∑\sum$ is the instruction to add.
$r$ is the index variable, which keeps track of the terms.
$r = 1$ is the lower limit, so the first value used is 1.
The number above the sigma is the upper limit, so the final value used is 5.
$2 r + 1$ is the summand, meaning the expression whose values are added.

To expand the sum, substitute each whole-number value of $r$ from the lower limit to the upper limit, including both limits.

Example

Expanding a sigma sum

Expand $∑r=15(2r+1)\displaystyle \sum_{r=1}^{5}(2r+1)$ .

The index runs through $r = 1, 2, 3, 4, 5$ , so substitute each of these values into $2 r + 1$ .
This gives $∑r=15(2r+1)=(2(1)+1)+(2(2)+1)+(2(3)+1)+(2(4)+1)+(2(5)+1).\begin{aligned} \sum_{r=1}^{5}(2r+1) &=(2(1)+1)+(2(2)+1)+(2(3)+1)\\ &\quad +(2(4)+1)+(2(5)+1). \end{aligned}$
Simplifying each term gives $3 + 5 + 7 + 9 + 11 = 35.$

Common Mistake

Leaving out an endpoint

Both limits are included. If the index runs from 1 to 5, there are five terms, not four.

Counting the number of terms

If the index runs through every integer from $a$ to $b$ , the number of terms is

b - a + 1.

For example, $∑r=410r\displaystyle \sum_{r=4}^{10}r$ contains $10 - 4 + 1 = 7$ terms.

Key Idea

Limits are values, not term counts

The upper limit gives the final value of the index. It does not necessarily give the number of terms unless the lower limit is 1.

Evaluating sums directly

For a short sum, the safest method is often to expand it and calculate each term.

Be especially careful when the summand contains a power or brackets. For example, in $r+1)^2$ , you must square the whole value of $r + 1$ .

Example

Evaluating a sum with squares

Evaluate $∑r=25(r+1)2\displaystyle \sum_{r=2}^{5}(r+1)^2$ .

The index values are $r = 2, 3, 4, 5$ , giving $2+1)^2+(3+1)^2+(4+1)^2+(5+1)^2.$
Evaluate each squared term: $3^2+4^2+5^2+6^2=9+16+25+36.$
Add the values to obtain $∑r=25(r+1)2=86.\sum_{r=2}^{5}(r+1)^2=86.$

Tip

Check the first and last terms

Before adding, substitute the lower limit and upper limit separately. This quickly checks that your expansion starts and finishes with the correct terms.

Writing a series using sigma notation

To convert an expanded series into sigma notation, you need to identify:

a formula that generates each term;
a suitable index variable;
the starting and finishing values of that index.

The index letter is a placeholder. Letters such as $r$ , $k$ and $i$ are commonly used, and changing the letter does not change the sum:

∑r=1nr2=∑k=1nk2.\sum_{r=1}^{n}r^2=\sum_{k=1}^{n}k^2.

Example

Writing an arithmetic series compactly

Write $5+8+11+⋯+355+8+11+\cdots+35$ using sigma notation.

The terms increase by 3, so they form an arithmetic sequence. Starting with index $r = 1$ , its rth term is $5 + 3 (r - 1) = 3 r + 2.$
Find the index of the final term by solving $3 r + 2 = 35,$ which gives $r = 11$ .
Therefore the series is $∑r=111(3r+2).\sum_{r=1}^{11}(3r+2).$

There may be more than one correct way to represent the same series. For example,

5+8+11+⋯+355+8+11+\cdots+35

can also be written as

∑r=010(3r+5).\sum_{r=0}^{10}(3r+5).

The limits and the summand have both changed, but the generated terms are identical.

Common Mistake

Changing limits without changing the formula

If you shift the starting value of the index, you usually need to adjust the summand as well. Always expand the first two terms and the last term to check your notation.

Algebra with sigma notation

Sigma notation follows the usual rules of addition and multiplication. Sums with the same limits can be combined:

∑r=1nar+∑r=1nbr=∑r=1n(ar+br).\sum_{r=1}^{n}a_r+\sum_{r=1}^{n}b_r = \sum_{r=1}^{n}(a_r+b_r).

A constant factor can be taken outside a sum:

∑r=1ncar=c∑r=1nar.\sum_{r=1}^{n}ca_r = c\sum_{r=1}^{n}a_r.

However, adding a constant inside a sum adds that constant once for every value of the index:

∑r=1nc=cn.\sum_{r=1}^{n}c=cn.

Consequently,

∑r=1n(ar+b)=a∑r=1nr+bn.\sum_{r=1}^{n}(ar+b) = a\sum_{r=1}^{n}r+bn.

Example

Splitting and evaluating a sum

Evaluate $∑r=120(3r−2)\displaystyle \sum_{r=1}^{20}(3r-2)$ .

Split the summand and take the constant factor outside: $∑r=120(3r−2)=3∑r=120r−∑r=1202.\sum_{r=1}^{20}(3r-2) = 3\sum_{r=1}^{20}r-\sum_{r=1}^{20}2.$
Use $∑r=120r=20(21)2=210\displaystyle \sum_{r=1}^{20}r=\frac{20(21)}{2}=210$ , and note that the constant 2 is added 20 times: $∑r=1202=2(20)=40.\sum_{r=1}^{20}2=2(20)=40.$
Therefore, $3 (210) - 40 = 590.$

Common Mistake

Treating a constant as a single term

In $∑r=1n(r+4)\displaystyle \sum_{r=1}^{n}(r+4)$ , the 4 occurs in every term. Its total contribution is $4 n$ , not 4.

Standard summation formulae

For large upper limits, expanding every term is inefficient. The following standard results are useful:

∑r=1n1=n,∑r=1nr=n(n+1)2,∑r=1nr2=n(n+1)(2n+1)6,∑r=1nr3=(n(n+1)2)2.\begin{aligned} \sum_{r=1}^{n}1&=n,\\ \sum_{r=1}^{n}r&=\frac{n(n+1)}{2},\\ \sum_{r=1}^{n}r^2&=\frac{n(n+1)(2n+1)}{6},\\ \sum_{r=1}^{n}r^3&=\left(\frac{n(n+1)}{2}\right)^2. \end{aligned}

These formulae apply directly when the lower limit is 1. If the lower limit is different, subtract the unwanted beginning of the sum.

For example,

∑r=4nr2=∑r=1nr2−∑r=13r2.\sum_{r=4}^{n}r^2 = \sum_{r=1}^{n}r^2-\sum_{r=1}^{3}r^2.

Example

Using a standard sum with a shifted lower limit

Evaluate $∑r=512r2\displaystyle \sum_{r=5}^{12}r^2$ .

Rewrite the required sum as the sum from 1 to 12 minus the sum from 1 to 4: $∑r=512r2=∑r=112r2−∑r=14r2.\sum_{r=5}^{12}r^2 = \sum_{r=1}^{12}r^2-\sum_{r=1}^{4}r^2.$
Apply the square-sum formula: $12(13)(25)6−4(5)(9)6.\frac{12(13)(25)}{6}-\frac{4(5)(9)}{6}.$
Calculate the difference: $650 - 30 = 620.$

Common Mistake

Check the required lower limit

To find a sum beginning at $r = 5$ , subtract the terms up to and including $r = 4$ . Subtracting up to $r = 5$ would remove one term too many.

Sigma notation and known series

Sigma notation can also describe arithmetic and geometric series. For example,

∑r=1n(a+(r−1)d)\sum_{r=1}^{n}\bigl(a+(r-1)d\bigr)

represents the first $n$ terms of an arithmetic sequence with first term $a$ and common difference $d$ .

Similarly,

r−1\sum_{r=1}^{n}ar^{\,r-1}

would be ambiguous because the same letter is being used as both the index and a fixed common ratio. It is clearer to use a different index:

∑k=1naqk−1,\sum_{k=1}^{n}aq^{k-1},

where $a$ is the first term and $q$ is the common ratio.

Key Idea

The index is a changing variable

Quantities such as $a$ , $d$ , $q$ and $n$ remain fixed while the index runs through its stated integer values.

Exam technique

In the exam

Read the lower limit, upper limit and summand separately before doing any calculation.
For an unfamiliar sum, write out its first two terms and final term to expose indexing errors.
Use standard summation formulae for large sums, but adjust carefully when the lower limit is not 1.
When constructing sigma notation, expand your answer briefly to check that it reproduces the original series.

Self review

Check yourself

Can you expand and evaluate $∑r=26(3r−1)\displaystyle \sum_{r=2}^{6}(3r-1)$ ?
How would you write $4+9+14+⋯+494+9+14+\cdots+49$ using sigma notation?
How could you evaluate $∑r=720r2\displaystyle \sum_{r=7}^{20}r^2$ without listing all fourteen terms?

1.7.3 Sigma notation (A-level only)

What you'll learn

From sequences to series

Understanding sigma notation

Counting the number of terms

Evaluating sums directly

Writing a series using sigma notation

Algebra with sigma notation

Standard summation formulae

Sigma notation and known series

Review 1.4.7 Sigma notation (A-level only) by teaching Genie

1.7.3 Sigma notation (A-level only)

What you'll learn

From sequences to series

Understanding sigma notation

Counting the number of terms

Evaluating sums directly

Writing a series using sigma notation

Algebra with sigma notation

Standard summation formulae

Sigma notation and known series

Review 1.4.7 Sigma notation (A-level only) by teaching Genie

1.7.3 Sigma notation (A-level only)

What you'll learn

From sequences to series

Understanding sigma notation

Counting the number of terms

Evaluating sums directly

Writing a series using sigma notation

Algebra with sigma notation

Standard summation formulae

Sigma notation and known series

Review 1.4.7 Sigma notation (A-level only) by teaching Genie

1.4.7 Sigma notation (A-level only) Revision Guide

1.7.3 Sigma notation (A-level only)

What you'll learn

From sequences to series

Understanding sigma notation

Counting the number of terms

Evaluating sums directly

Writing a series using sigma notation

Algebra with sigma notation

Standard summation formulae

Sigma notation and known series

Review 1.4.7 Sigma notation (A-level only) by teaching Genie

1.4.7 Sigma notation (A-level only) Revision Guide