Standard Series (Edexcel International AS Further Maths)

Revision Note

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 10 April 2024

Sums of Natural Numbers

What is sigma notation?

Sigma notation represents sums as follows: $\sum_{r = 1}^{5} r = 1 + 2 + 3 + 4 + 5$
- $r$ counts in integers from the lower limit to the upper limit
It works for functions of $r$
- $\sum_{r = 1}^{5} (3 r - 1) = (3 \times 1 - 1) + (3 \times 2 - 1) + (3 \times 3 - 1) + . . . + (3 \times 5 - 1)$
The sum of the first $n$ terms has an upper limit of $n$
- $\sum_{r = 1}^{n} r = 1 + 2 + 3 + . . . + n$

How do I write a series as the difference of two sums?

When the lower limit is not 1, such as $\sum_{r = 5}^{10} r = 5 + 6 + 7 + . . . + 10$ , you can write it as the difference between two sums:
- $\sum_{r = 5}^{10} r = \sum_{r = 1}^{10} r - \sum_{r = 1}^{4} r$
- This is because $5 + 6 + 7 + 8 + 9 + 10 = (1 + 2 + 3 + 4 + . . . + 10) - (1 + 2 + 3 + 4)$
  - The 1, 2, 3 and 4 cancel out, leaving 5, 6, ..., 10
- Be careful: the upper limit of the second sum is one less than the lower limit of the original sum!

What properties of sigma notation do I need to know?

You need to know the following rules:
- $\sum_{r = 1}^{n} k = k + k + . . . + k = k n$
  - The sum of a constant is the constant $\times n$
- $\sum_{r = 1}^{n} (a r + b) = a \sum_{r = 1}^{n} r + b n$
  - Coefficients of $r$ can come out of the sum
  - Constants are multiplied by $n$

What is the formula for the sum of the natural numbers?

$\sum_{r = 1}^{n} r = 1 + 2 + 3 + . . . + n$ is the sum of the natural numbers
- Natural numbers are positive integers
It has the standard formula:
- $\sum_{r = 1}^{n} r = \frac{1}{2} n (n + 1)$
- You can work the formula out using arithmetic series
  - first term 1, common difference 1
You may use the formula in calculations without proof

Examiner Tips and Tricks

The formula for the sum of natural numbers is not given in the Formulae Booklet!

Worked Example

A sum of $n$ terms is given by $\sum_{r = 1}^{n} (2 r + 1)$ .

(a) Using any standard summation formulae, show that $\sum_{r = 1}^{n} (2 r + 1) = n (n + A)$ where $A$ is a positive integer to be found.

Use the rule that $\sum_{r = 1}^{n} (a r + b) = a \sum_{r = 1}^{n} r + b n$

$\sum_{r = 1}^{n} (2 r + 1) = 2 \sum_{r = 1}^{n} r + n$

Substitute in the formula $\sum_{r = 1}^{n} r = \frac{1}{2} n (n + 1)$

$\sum_{r = 1}^{n} (2 r + 1) = 2 (\frac{1}{2} n (n + 1)) + n$

Expand and simplify

$\begin{array}{rcl} \sum_{r = 1}^{n} (2 r + 1) & = & n (n + 1) + n \\ = & n^{2} + 2 n \end{array}$

Factorise the right-hand side

$\begin{array}{rcl} \sum_{r = 1}^{n} (2 r + 1) & = & n (n + 2) \end{array}$

Check it has the right form, $n (n + A)$

$\begin{array}{rcl} \sum_{r = 1}^{n} (2 r + 1) & = & n (n + 2) \end{array}$ where $\begin{array}{rcl} A & = & 2 \end{array}$

(b) Hence find the sum of the odd numbers from 3 to 81.

It often helps to write out the first few terms and the last term of the series

$\sum_{r = 1}^{n} (2 r + 1) = (2 \times 1 + 1) + (2 \times 2 + 1) + (2 \times 3 + 1) + . . . + (2 \times n + 1)$

Simplify these terms

$\sum_{r = 1}^{n} (2 r + 1) = 3 + 5 + 7 + . . . + (2 n + 1)$

This is the sum of the odd numbers from 3 to $(2 n + 1)$
The question wants the sum of odd numbers from 3 to 81
Use the last term to find $n$

$\begin{array}{rcl} 2 n + 1 & = & 81 \\ 2 n & = & 80 \\ n & = & 40 \end{array}$

The sum of the first 40 terms gives the sum of odd numbers from 3 to 81
Substitute $n = 40$ into the formula in part (a)

$40 \times (40 + 2)$

1680

Sums of Squares

What is the formula for the sum of the squares of the natural numbers?

$\sum_{r = 1}^{n} r^{2} = 1^{2} + 2^{2} + 3^{2} + . . . + n^{2} = 1 + 4 + 9 + . . . + n^{2}$ is the sum of the squares of the natural numbers
It has it the standard formula:
- $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$
- You may use the formula in calculations without proof
Note that $\sum_{r = 1}^{n} (a r^{2} + b) = a \sum_{r = 1}^{n} r^{2} + b n$
- Coefficients of $r^{2}$ can come out of the sum
- Constants are multiplied by $n$

Examiner Tips and Tricks

This formula is given in the Formulae Booklet.

Worked Example

(a) Evaluate $\sum_{r = 1}^{30} r^{2}$

Evaluate means "find the value of"
This is the sum of the first 30 square numbers
Substitute $n = 30$ into the formula $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$

$\begin{array}{rcl} \sum_{r = 1}^{30} r^{2} & = & \frac{1}{6} \times 30 \times (30 + 1) (2 \times 30 + 1) \\ = & \frac{1}{6} \times 30 \times 31 \times 61 \end{array}$

9455

(b) Show that $\sum_{r = m + 1}^{2 m} r^{2} = \frac{1}{6} m (P m + 1) (Q m + 1)$ where $P < Q$ and where $P$ and $Q$ are prime numbers to be found.

The lower limit is not 1, so write it as the difference between two sums
It can help to imagine simpler numbers, $\sum_{r = 5}^{10} r^{2} = \sum_{r = 1}^{10} r^{2} - \sum_{r = 1}^{4} r^{2}$