Introduction
• Chapter 5 of FP1 focuses on methods to
calculate the sum of a series of numbers
• The main focus is based around summing the
first ‘n’ natural numbers of a given power
• You will also become familiar with the proper
series notation (you may already have seen this
if you have covered Arithmetic and Geometric
sequences from C1 and C2)
Series
You need to be able to understand and use the
Σ notation
This is the last value you put
in to the formula (hence giving
you the last term to be
summed)
This
means
‘the sum
of’
10
5𝑟 + 1
𝑟=1
This is the first value you put
in to the formula (hence
giving you the first term to
be summed)
This is the last value you put
in to the formula (hence giving
you the last term to be
summed)
This is the
formula for the
sequence you are
calculating the
sum of (the ‘nth’
term)
This
means
‘the sum
of’
40
𝑟2
𝑟=0
This is the
formula for the
sequence you are
calculating the
sum of (the ‘nth’
term)
This is the first value you put
in to the formula (hence
giving you the first term to
be summed)
5A
Series
You need to be able to understand and use the
Σ notation
This is the last value you put
in to the formula (hence giving
you the last term to be
summed)
This
means
‘the sum
of’
20
𝑟 2 − 3𝑟
𝑟=10
This is the last value you put
in to the formula (hence giving
you the last term to be
summed)
This is the
formula for the
sequence you are
calculating the
sum of (the ‘nth’
term)
This is the first value you put
in to the formula (hence
giving you the first term to
be summed)
In this example you are effectively calculating
the sum of the terms from the 10th to the 20th
This
means
‘the sum
of’
𝑛
4𝑟 − 5
𝑟=1
This is the
formula for the
sequence you are
calculating the
sum of (the ‘nth’
term)
This is the first value you put
in to the formula (hence
giving you the first term to
be summed)
In this example you are not told how many
terms there are, so the answer will be in terms
of ‘n’, the number of terms…
5A
Series
You need to be able to understand and use the
Σ notation
Write out the terms defined by the following
notation, and hence calculate the sum of the
series:
10
(2𝑟 − 1)
𝑟=1
Put r = 1 in for the first term, r = 2 in for
the second and so on…
Stop after calculating the term for r = 10
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19
Write down the first 3 terms of, and the final
term of, the sequence indicated by the notation
below
𝑛
(2𝑟 + 1)
𝑟=3
Put r = 3 in for the first term, r = 4 in for
the second and so on…
Put ‘n’ in for the final term…
7 + 9 + 11 + ……………………………………+ (2n + 1)
= 100
5A
Series
You need to be able to understand and use
the Σ notation
Write the sequence below using the ∑
notation.
8
(3𝑟 − 1)
𝑟=1
2 + 5 + 8 + 11 + 14 + 17 + 20 + 23
 You need the formula for the sequence
and the values to put in for the first and
final terms!
 Formula for the sequence = ‘3n – 1’ (from
GCSE maths!
 Substituting 1 in will get 2 – the first
term
 Substituting 8 in will get 23 – the final
term
It is usually possible to do this several ways
 The notation below will give the same
sequence (although is not as easy to work
out!)
9
(3𝑟 − 4)
𝑟=2
5A
Series
𝑛−2
You need to be able to understand and use
the Σ notation
𝑛(𝑛 − 1)
Write the sequence below using the ∑
notation.
𝑟=1
1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + …… + (n – 1)(n – 2)
 You need the formula for the sequence
and the values to put in for the first and
final terms!
 Formula for the sequence = n(n+1)
(multiply the term by the number one
bigger than it!
 Substituting 1 in will get the first term
 To get the final term, we sub in (n – 2)
𝑛(𝑛 + 1)
(𝑛 − 2)(𝑛 − 2 + 1)
= (𝑛 − 2)(𝑛 − 1)
= (𝑛 − 1)(𝑛 − 2)
Replace n with ‘n – 2’
Simplify
Rewrite
 This is the final term!
5A
Series
You need to be able to use the formula
for the sum of the first n natural numbers
The sum of the first ‘n’ natural numbers:
1 + 2 + 3 + 4 + …… + n
𝑛
is an arithmetic sequence with ‘n’ terms, with
a = 1 and l (last term) = n
𝑟=1
𝑛
𝑟 = 𝑛+1
2
The formula for the sum of the first ‘n’
natural numbers is shown to the right:
You DO NOT get
given this formula
on the exam!
This is just saying the sum of the first ‘n’
natural numbers
This is the formula for the sum
of the first ‘n’ natural numbers
 The formula for the sequence is just ‘r’
as when you substitute in the term
number, that is the term itself!
 For example if you wanted
the sum of the first 30
numbers, let n = 30 and
calculate the answer!
5B
𝑛
𝑟=
Series
You need to be able to use the formula
for the sum of the first n natural numbers
𝑟=1
𝑛
𝑟=
𝑟=1
Calculate the sum of the series indicated
below:
50
𝑟=
50
𝑟=1
𝑟
𝑛
𝑛+1
2
50
50 + 1
2
We want the sum
of the first 50
terms, so n = 50
Calculate
50
𝑟=1
𝑛
𝑛+1
2
𝑟 = 1275
𝑟=1
𝑛
𝑛+1
2
𝑛(𝑛 + 1)
2
It is fine (and sometimes
easier) to use the formula
in this form!
5B
𝑛
Series
𝑟=1
60
𝑟
𝑟=21
𝑟=
𝑟=1
𝑛
𝑛+1
2
The sum of the numbers
from 21 to 60…
You need to be able to use the formula
for the sum of the first n natural numbers
Calculate the sum of the series indicated
below:
𝑛
𝑛(𝑛 + 1)
𝑟=
2
0
10
20
30
40
50
60
… Will be equal to the sum of the numbers
from 1 to 60, subtract the numbers from 1
to 20…
The notation will look like this…
60
This is asking you to find the sum of the
numbers from 21 to 60
60
𝑟 =
𝑟=21
Sum from
21 to 60
20
𝑟 −
𝑟=1
Sum from
1 to 60
𝑟
Notice the number
here will always be
one less than the
one at the start!
𝑟=1
Sum from
1 to 20
5B
𝑛
Series
You need to be able to use the formula
for the sum of the first n natural numbers
Calculate the sum of the series indicated
below:
60
60
𝑟 =
𝑟=21
𝑟=
𝑟=1
𝑛
𝑛+1
2
𝑟 −
𝑟
𝑟=1
Sub in values
for each part
𝑟 =
𝑟=21
𝑛
20
𝑟=1
60
60
𝑟
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
60 × 61
20 × 21
−
2
2
Calculate
𝑟=21
= 1620
This is asking you to find the sum of the
numbers from 21 to 60
So the sum of the numbers from
21 to 60 is 1620!
5B
𝑛
Series
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑟=
𝑟=1
𝑛
𝑛+1
2
You need to be able to use the formula
for the sum of the first n natural numbers
Remember the link between the
starting number and the sum we
subtract!
This is the general form for the problem you
have seen (where you sum the numbers of a
section of natural numbers, not starting on 1)
𝑛
𝑛
𝑓(𝑟) =
𝑟=𝑘
𝑘−1
𝑓(𝑟) −
𝑟=1
The sum of the
numbers from
1 to n
𝑓(𝑟)
𝑟=1
The sum of the
numbers we will be
removing (k – 1)
5B
𝑛
Series
2𝑁−1
𝑘 = 2𝑁 2 − 𝑁 − 10
𝑁≥3
𝑟=
𝑟=1
𝑘 −
𝑘=5
𝑘=1
𝑘
𝑘=1
Write out the
formula twice as you
will need it for both!
=
𝑛(𝑛 + 1)
𝑛(𝑛 + 1)
−
2
2
This type of question can look confusing but
in reality you proceed as before
=
(2𝑁 − 1)(2𝑁 − 1 + 1) 4 × 5
−
2
2
 We want the sum of the natural numbers
from 5 to 2N – 1
=
20
(2𝑁 − 1)(2𝑁)
−
2
2
𝑘=5
 This will be the sum of the natural
numbers from 1 to 2N – 1, subtract the
numbers from 1 to 4.
𝑛
𝑛+1
2
4
2𝑁−1
𝑘 =
2𝑁−1
𝑛
Write out the formula for the
numbers you want…
You need to be able to use the formula
for the sum of the first n natural numbers
Show that:
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
4𝑁 2 − 2𝑁 − 20
=
2
Sub (2N – 1)
in for the
1st, and 4 in
for the 2nd
Simplify or
calculate
where
possible
Expand brackets
and group up
Simplify
= 2𝑁 2 − 𝑁 − 10
5B
𝑛
Series
You need to be able to use the formula
for the sum of the first n natural numbers
Show that:
2𝑁−1
𝑘 = 2𝑁 2 − 𝑁 − 10
𝑁≥3
𝑘=5
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑟=
𝑟=1
𝑛
𝑛+1
2
But why N ≥ 3?
 Remember that you are told k = 5, meaning
the first number in the sequence we are
summing should be 5!
 If you use the value of N = 3, the sum of the
sequence is 5, hence 3 is the lowest number
we can put in!
This type of question can look confusing but
in reality you proceed as before
 We want the sum of the natural numbers
from 5 to 2N – 1
 This will be the sum of the natural
numbers from 1 to 2N – 1, subtract the
numbers from 1 to 4.
5B
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑟=
𝑟=1
𝑛
𝑛+1
2
Series
You need to be able to split up parts of a
sequence and sum them separately
You can split up series sums of the form:
𝑛
(𝑎𝑟 + 𝑏)
𝑟=1
into 2 separate ‘series sums’ as follows:
𝑛
𝑎
𝑛
𝑟 + 𝑏
𝑟=1
1
𝑟=1
This allows you to then use the sum formulae
for the sequence overall!
5C
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑟=
𝑟=1
𝑛
𝑛+1
2
Series
𝑛
You need to be able to split up parts
of a sequence and sum them
separately
= 3 × 1 + 2 + 3× 2 + 2 + 3× 3 + 2 + 3 × 4 + 2 ………
𝑛
(3𝑟 + 2)
This is equal to the sum of the multiplied terms, added to the sum of the 2s
𝑟=1
= 3 × 1 + 3 × 2 + 3 × 3 + 3 × 4 + 2+ 2 + 2 + 2 ……
Can be written as:
𝑛
𝑛
𝑟 + 2
𝑟=1
𝑟=1
If we wrote out the first few terms of this sequence…
Show that:
3
(3𝑟 + 2)
1
𝑟=1
We can ‘factorise’ the 3 out of the multiplied terms and factorise
a 2 from the added terms…
= 3 × 1 + 2 + 3 + 4 ……… + 2 1 + 1 + 1 + 1 ………
This is 3 multiplied by the sum
of the first ‘n’ numbers
represented by the formula ‘r’
This is 2 multiplied by ‘n’ 1s
𝑛
3
𝑛
𝑟 + 2
𝑟=1
1
𝑟=1
5C
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑟=
𝑟=1
𝑛
𝑛+1
2
Series
25
You need to be able to split up parts
of a sequence and sum them
separately
(3𝑟 + 1)
Split into two
separate parts as
you have seen
𝑟=1
25
Evaluate:
3
25
𝑟 +
𝑟=1
(3𝑟 + 1)
𝑟=1
=
You need to split this up and sum the
parts separately!
=
25
1
𝑟=1
3𝑛(𝑛 + 1)
+ 𝑛(1)
2
3(25)(25 + 1)
+ 25(1)
2
Write the formulae for the
sums. Remember the 3 at the
start of the first one!
 We will also have ‘n’ lots of 1
Sub in n = 25 (25 terms
to add up)
Calculate
= 1000
So the first 25 terms of the sequence with the
formula (3r + 1) will add up to 1000!
5C
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑟=
𝑟=1
𝑛
𝑛+1
2
Series
𝑛
You need to be able to split up parts
of a sequence and sum them
separately
(7𝑟 − 4)
𝑟=1
𝑛
Show that:
𝑛
𝑛
7𝑛 − 1
(7𝑟 − 4) =
2
𝑟=1
7
=
The sum of the
first ‘n’ terms of
this sequence
Is given by this
formula, where
‘n’ is the number
of terms
The two expressions are
equivalent!
𝑟 −
4
𝑟=1
7𝑛(𝑛 + 1)
− 4𝑛
2
8𝑛
7𝑛(𝑛 + 1)
−
2
2
7𝑛 𝑛 + 1 − 8𝑛
=
2
=
In this case you should proceed as
normal, but use ‘n’ instead!
𝑛
𝑟=1
=
Split up as two
separate sums
7𝑛2 + 7𝑛 − 8𝑛
2
7𝑛2 − 𝑛
=
2
𝑛(7𝑛 − 1)
=
2
Remember the 7 on the
first expression!
 We also have n lots of 4
Write ‘4n’ as fraction over
2 (for grouping)
Group terms
Expand the
bracket
Group terms
Factorise
5C
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑟=
𝑟=1
𝑛
𝑛+1
2
Series
Write as one sum subtract another
You need to be able to split up parts
of a sequence and sum them
separately
Show that:
𝑛
(7𝑟 − 4) =
𝑟=1
𝑛
7𝑛 − 1
2
Hence, calculate the value of:
50
(7𝑟 − 4)
50
19
50
(7𝑟 − 4) =
𝑟=20
(7𝑟 − 4) −
𝑟=1
(7𝑟 − 4)
𝑟=1
=
𝑛(7𝑛 − 1)
𝑛(7𝑛 − 1)
−
2
2
=
50(7(50) − 1)
19(7(19) − 1)
−
2
2
= 8725 − 1254
= 7471
Write the formula
separately for
each sum
Sub 50 into
the first and
19 into the
second
Calculate
each
Finish
off!
𝑟=20
Here, you can use the formula you’re
given – remember that this will be the
sum of the first 50 terms subtract the
sum of the first 19!
5C
Series
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
The sum of a sequence of squared
numbers is given as follows:
𝑛
𝑟2 =
𝑟=1
𝑛(𝑛 + 1)(2𝑛 + 1)
6
𝑜𝑟
𝑛
(𝑛 + 1)(2𝑛 + 1)
6
And the formula for the sum of a
sequence of cubes is:
𝑛
𝑛2 𝑛 + 1
𝑟 =
4
𝑟=1
3
2
𝑜𝑟
𝑛2
𝑛+1
4
You will see where these come from in
chapter 6!
2
You get given these formulae in
these forms in the exam booklet!
(Remember you do not get the
formula for a linear sequence!)
5D
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑟=1
𝑛2 𝑛 + 1
4
2
30
𝑟2
𝑟=1
Write out the formula for
a squared sequence
=
𝑛(𝑛 + 1)(2𝑛 + 1)
6
=
(30)(30 + 1)(2(30) + 1)
6
=
(30)(31)(61)
6
30
𝑟2
𝑟3 =
Series
𝑟=1
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
Evaluate:
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
2
𝑟=1
Sub in n = 30 as
we want 30
terms
Simplify the
numerator (if
necessary!)
Calculate
= 9455
5D
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
2
𝑟3 =
Series
𝑟=1
𝑟=1
40
40
𝑟3 =
19
𝑟3 −
𝑟=1
𝑟3
𝑟=1
Evaluate:
𝑟3
𝑛2 𝑛 + 1
=
4
𝑟=20
Remember for this one you need the sum
of the first 40 terms, subtract the first
19 terms!
2
Write it as one sum subtract another
𝑟=20
40
𝑛2 𝑛 + 1
4
=
2
𝑛2 𝑛 + 1
−
4
(40)2 40 + 1
4
2
−
2
(19)2 19 + 1
4
Write out the formula
for the cubed
sequence twice
2
Sub in 40 for
the first and
19 for the
second
Calculate
= 672400 − 36100
Finish the sum!
= 636300
5D
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
2𝑛
You need to be able to calculate the
𝑟2
sum of a sequence based on powers of
𝑟=𝑛+1
2 and 3
𝑟=𝑛+1
𝑛2 𝑛 + 1
4
2
𝑛
𝑟2 −
=
𝑟=1
𝑟2
Write out the
formula twice
𝑟=1
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑛(𝑛 + 1)(2𝑛 + 1)
−
6
6
=
2𝑛(2𝑛 + 1)(4𝑛 + 1)
𝑛(𝑛 + 1)(2𝑛 + 1)
−
6
6
This one is more algebraic but you still
approach it the same way!
=
The first value we put in the sequence
will be ‘n + 1’
= 𝑛(2𝑛 + 1) 2 4𝑛 + 1 − (𝑛 + 1)
6
The final value we put in will be ‘2n’
= 𝑛(2𝑛 + 1) 8𝑛 + 2 − 𝑛 − 1
6
So we want the sum of the first ‘2n’
terms, subtract the first ‘n’ terms (same
as if we were using numbers!)
𝑟=1
=
2𝑛
𝑟2
𝑟3 =
Series
𝑟=1
2𝑛
Find:
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
2
2𝑛 2𝑛 + 1 4𝑛 + 1 − 𝑛(𝑛 + 1)(2𝑛 + 1)
6
[
[
= 𝑛(2𝑛 + 1)(7𝑛 + 1)
6
]
]
Sub ‘2n’ into the
first and ‘n’ into the
second
You can write this
as one fraction
This is the key step –
you can factorise as
n(2n+1) is common to
both terms!
Expand the terms in
the square bracket
Simplify the square
bracket (which can now
be written as a ‘normal’
bracket!)
The factorising step is crucial here – otherwise you will end up
trying to factorise a cubic which can take a long time!
5D
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
Series
𝑟=1
2𝑛
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
𝑟2 =
𝑟=𝑛+1
𝑟3 =
𝑟=1
𝑛2 𝑛 + 1
4
2
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
If n = 1
Find:
2𝑛
2
𝑟2
𝑟2
𝑟=𝑛+1
=
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
Verify that the result is correct for n =
1, 2 and 3
(This can show the formula is working,
although in reality isn’t a proof in itself!)
𝑟=2
The first number we
put in is 2, which is
also the last number
we put in
Sequence 
4
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
=
1(2 + 1)(7 + 1)
6
=
24
6
=4
So the numbers in the
sequence just add up to 4!
 Let’s check the
formula!
5D
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
Series
𝑟=1
2𝑛
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
𝑟2 =
𝑟=𝑛+1
𝑟3 =
𝑟=1
𝑛2 𝑛 + 1
4
2
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
If n = 2
Find:
2𝑛
4
𝑟2
𝑟2
𝑟=𝑛+1
=
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
Verify that the result is correct for n =
1, 2 and 3
(This can show the formula is working,
although in reality isn’t a proof in itself!)
𝑟=3
The first number we
put in is 3, and the
last number we put in
in 4
Sequence  9, 16
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
=
2(4 + 1)(14 + 1)
6
=
150
6
= 25
So the numbers in the
sequence add up to 25
 Let’s check the
formula!
5D
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
2
𝑟3 =
Series
𝑟=1
2𝑛
You need to be able to calculate the
sum of a sequence based on powers of
2 and 3
𝑟2 =
𝑟=𝑛+1
𝑟=1
𝑛2 𝑛 + 1
4
2
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
If n = 3
Find:
2𝑛
6
𝑟2
𝑟2
𝑟=𝑛+1
=
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
Verify that the result is correct for n =
1, 2 and 3
(This can show the formula is working,
although in reality isn’t a proof in itself!)
𝑟=4
The first number we
put in is 4, and the
last number we put in
in 6
Sequence  16, 25, 36
So the numbers in the
sequence add up to 77
 Let’s check the
formula!
𝑛(2𝑛 + 1)(7𝑛 + 1)
6
=
3(6 + 1)(21 + 1)
6
=
462
6
= 77
So the formula seems
to be working fine!
5D
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
𝑛2 𝑛 + 1
4
2
Series
𝑟=1
You need to be able to use all you have learnt
to calculate the sum of a more complex series,
made up of several terms
As you saw in section 5C, you can take out a
coefficient of a term in order to sum it.
You can also do this with the sums for r2 and r3.
 For example:
𝑛
𝑛
4𝑟 2 = 4
𝑟=1
𝑟2
𝑟=1
=
4𝑛(𝑛 + 1)(2𝑛 + 1)
6
 You need to remember to include the
coefficient in the formula though!
5E
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
𝑛2 𝑛 + 1
4
2
Series
𝑟=1
You need to be able to use all you have learnt
to calculate the sum of a more complex series,
made up of several terms
𝑛
𝑟2 + 𝑟 − 2
𝑟=1
𝑛
𝑛
2
Show that:
𝑛
𝑟2 + 𝑟 − 2 =
𝑟=1
=
=
=
𝑛
(𝑛 + 4)(𝑛 − 1)
3
2𝑛(𝑛2 + 3𝑛 − 4)
6
𝑛 (𝑛2 + 3𝑛 − 4)
3
𝑛 (𝑛 + 4)(𝑛 − 1)
3
Divide numerator and
denominator by 2
𝑟 +
𝑛
𝑟 −
𝑟=1
Write as
separate sums
𝑟=1
2
𝑟=1
=
𝑛(𝑛 + 1)
𝑛(𝑛 + 1)(2𝑛 + 1)
− 2𝑛
+
2
6
=
12𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
3𝑛(𝑛 + 1)
−
+
6
6
6
𝑛 𝑛 + 1 2𝑛 + 1 + 3𝑛 𝑛 + 1 − 12𝑛
=
6
=
Factorise!
=
=
=
[
𝑛 (𝑛 + 1)(2𝑛 + 1) + 3(𝑛 + 1) − 12
]
6
[
2
𝑛 2𝑛 + 3𝑛 + 1 + 3𝑛 + 3 − 12
6
𝑛(2𝑛2 + 6𝑛 − 8)
6
2𝑛 (𝑛2 + 3𝑛 − 4)
6
]
Write the formula
for each part in
terms of n
Write all with a
common
denominator
Group up
‘Clever
factorisation’
Expand
brackets
Group terms
Take the factor 2
out of the bracket
5E
𝑛
𝑟=1
𝑛(𝑛 + 1)
𝑟=
2
𝑛
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
𝑛2 𝑛 + 1
4
2
Series
𝑟=1
𝑛
You need to be able to use all you have learnt
to calculate the sum of a more complex series,
made up of several terms
𝑟2 + 𝑟 − 2
𝑟=1
𝑟=1
1
2
+ 1 −2
=0
𝑟=2
2
2
+ 2 −2
=4
𝑟=3
3
2
+ 3 −2
= 10
4 + 10 + 18 + 28 + 40 … … … + 418
𝑟=4
4
2
+ 4 −2
= 18
You can see that this formula gives us the
sequence we are trying to find the sum of!
𝑟=?
Show that:
𝑛
𝑟2 + 𝑟 − 2 =
𝑟=1
𝑛
(𝑛 + 4)(𝑛 − 1)
3
Hence, calculate the sum of the series:
(The 0 at the start will not affect the sum
so can be ignored!)
𝑟2 + 𝑟 − 2
= 418
We need to know how many terms there
are, so have to find the value for r which
gives a term with a value of 418…
5E
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
2
Series
𝑟=1
You need to be able to use all you have learnt
to calculate the sum of a more complex series,
made up of several terms
Show that:
𝑛
𝑛2 𝑛 + 1
4
𝑟 2 + 𝑟 − 2 = 418
𝑟 + 𝑟 − 420 = 0
(𝑟 − 20)(𝑟 + 21) = 0
𝑛
𝑟 + 𝑟 − 2 = (𝑛 + 4)(𝑛 − 1)
3
2
𝑟=1
Hence, calculate the sum of the series:
4 + 10 + 18 + 28 + 40 … … … + 418
Subtract 418
2
Factorise
2 answers, only 1 is
possible though!
𝑟 = 20 𝑜𝑟 − 21
So we are finding the sum of the first 20 terms of
the sequence!
20
𝑟2 + 𝑟 − 2
𝑟=1
=
𝑛
(𝑛 + 4)(𝑛 − 1)
3
20
=
(20 + 4)(20 − 1)
3
= 3040
We can use the
formula we were given!
Sub in n = 20
Calculate
5E
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
𝑛2 𝑛 + 1
4
2
Series
𝑟=1
2
You need to be able to use all you have learnt = 2𝑛 𝑛 + 1
4
to calculate the sum of a more complex series,
made up of several terms
2
3𝑛2 𝑛 + 1
=
6
2
Find a formula for the sum of the series:
+
3𝑛(𝑛 + 1)
5𝑛(𝑛 + 1)(2𝑛 + 1)
−
2
6
+
5𝑛(𝑛 + 1)(2𝑛 + 1)
9𝑛(𝑛 + 1)
−
6
6
Combine
𝑛
𝑟(𝑟 + 3)(2𝑟 − 1)
𝑟=1
Expand the
bracket
𝑛
𝑟(2𝑟 + 5𝑟 − 3)
𝑟=1
Expand the
bracket again
𝑛
2𝑟 3 + 5𝑟 2 − 3𝑟
𝑟=1
𝑛
𝑛
2
𝑟 + 5
𝑟=1
=
2𝑛2 𝑛 + 1
4
𝑟
𝑟=1
2
+
Write as 3
separate sums
𝑛
3
2
=
=
2
− 3
𝑟
𝑟=1
3𝑛(𝑛 + 1)
5𝑛(𝑛 + 1)(2𝑛 + 1)
−
2
6
Write with
the same
denominator
=
=
3𝑛2 𝑛 + 1
2
+ 5𝑛 𝑛 + 1 2𝑛 + 1 − 9𝑛(𝑛 + 1)
6
[
𝑛(𝑛 + 1) 3𝑛(𝑛 + 1) + 5(2𝑛 + 1)− 9
]
6
[
2
𝑛(𝑛 + 1) 3𝑛 + 3𝑛 + 10𝑛 + 5 − 9
6
𝑛(𝑛 + 1)(3𝑛2 + 13𝑛 − 4)
6
Write using
the formulae
above.
Remember to
include the
coefficients!
]
‘Clever
factorisation’
Expand the inner
brackets
Simplify (you
should also
factorise if
possible)
5E
𝑛
𝑟=1
𝑛
𝑛(𝑛 + 1)
𝑟=
2
𝑛(𝑛 + 1)(2𝑛 + 1)
𝑟 =
6
𝑛
2
𝑟=1
𝑟3 =
Find a formula for the sum of the series:
Write as one sum subtract another
40
40
𝑟(𝑟 + 3)(2𝑟 − 1) =
𝑟=11
𝑟(𝑟 + 3)(2𝑟 − 1)
𝑟=1
𝑟=11
𝑟(𝑟 + 3)(2𝑟 − 1) −
𝑟=1
𝑟(𝑟 + 3)(2𝑟 − 1)
𝑟=1
=
𝑛(𝑛 + 1)(3𝑛2 + 13𝑛 − 4)
𝑛(𝑛 + 1)(3𝑛2 + 13𝑛 − 4)
−
6
6
=
10(11)(426)
40(41)(5316)
−
6
6
40
𝑟(𝑟 + 3)(2𝑟 − 1)
10
Write the formulae out twice, one for each sum!
𝑛
Hence, calculate the following:
2
Series
𝑟=1
You need to be able to use all you have learnt
to calculate the sum of a more complex series,
made up of several terms
𝑛(𝑛 + 1)(3𝑛2 + 13𝑛 − 4)
=
6
𝑛2 𝑛 + 1
4
Sub in
40 and
10
Calculate!
= 1445230
5E
Summary
• We have seen how to calculate the sum
of a series in various circumstances
• We have practiced the correct series
notation
• We have also seen and used the ‘clever
factorisation’ method for simplifying
expressions!