Standard Series (Edexcel International A Level Further Maths)

Revision Note

Author

Mark Curtis

Expertise

Maths

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

Exam Tip

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 $n$

Exam Tip

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, $\sum_{r = 5}^{10} r^{2} = \sum_{r = 1}^{10} r^{2} - \sum_{r = 1}^{4} r^{2}$

$\sum_{r = m + 1}^{2 m} r^{2} = \sum_{r = 1}^{2 m} r^{2} - \sum_{r = 1}^{m} r^{2}$

Now use the formula $\sum_{r = 1}^{n} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ for each sum on the right

$\sum_{r = m + 1}^{2 m} r^{2} = \frac{1}{6} (2 m) (2 m + 1) (4 m + 1) - \frac{1}{6} m (m + 1) (2 m + 1)$

Factorise out $\frac{1}{6}$ , $m$ and $(2 m + 1)$ then simplify

$\begin{array}{rcl} \sum_{r = m + 1}^{2 m} r^{2} & = & \frac{1}{6} m (2 m + 1) [2 (4 m + 1) - (m + 1)] \\ = & \frac{1}{6} m (2 m + 1) (7 m + 1) \end{array}$

Check this is in the right form
2 and 7 are prime numbers and 2 < 7

$\begin{array}{rcl} \sum_{r = m + 1}^{2 m} r^{2} & = & \frac{1}{6} m (2 m + 1) (7 m + 1) \end{array}$ where $\begin{array}{rcl} P & = & 2 \end{array}$ and $\begin{array}{rcl} Q & = & 7 \end{array}$

Sums of Cubes

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

$\sum_{r = 1}^{n} r^{3} = 1^{3} + 2^{3} + 3^{3} + . . . + n^{3} = 1 + 8 + 27 + . . . + n^{3}$ is the sum of the cubes of the natural numbers
It has it the standard formula:
- $\sum_{r = 1}^{n} r^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}$
- You may use the formula in calculations without proof
- It is the square of the formula for the sum of the natural numbers
  - ${(\sum_{r = 1}^{n} r)}^{2} = {(\frac{1}{2} n (n + 1))}^{2} = \frac{1}{4} n^{2} {(n + 1)}^{2}$
Note that $\sum_{r = 1}^{n} (a r^{3} + b) = a \sum_{r = 1}^{n} r^{3} + b n$
- Coefficients of $r^{3}$ can come out of the sum
- Constants are multiplied by $n$

Exam Tip

This formula is given in the Formulae Booklet.

Worked Example

(a) Find $1^{3} + 2^{3} + 3^{3} + . . . + 24^{3}$

This is the sum of the first 24 cube numbers
Substitute $n = 24$ into the formula $\sum_{r = 1}^{n} r^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}$

$\begin{array}{rcl} \sum_{r = 1}^{24} r^{3} & = & \frac{1}{4} \times 24^{2} \times {(24 + 1)}^{2} \\ = & \frac{1}{4} \times 24^{2} \times 25^{2} \end{array}$

90000

(b) Find and simplify a formula for the sum of the first $n$ even cube numbers.

The sum of even cube numbers is 2³ + 4³ + 6³ + ...
The first 5 would end on 10³
The first $n$ would end on ${(2 n)}^{3}$

$2^{3} + 4^{3} + 6^{3} + . . . + {(2 n)}^{3}$

The standard formula is for $\sum_{r = 1}^{n} r^{3}$
To get just the even numbers, replace $r$ with $2 r$

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

Cube the expression and use $\sum_{r = 1}^{n} a r^{3} = a \sum_{r = 1}^{n} r^{3}$

$\begin{array}{rcl} \sum_{r = 1}^{n} {(2 r)}^{3} & = & \sum_{r = 1}^{n} 8 r^{3} \\ = & 8 \sum_{r = 1}^{n} r^{3} \end{array}$

Now substitute in the standard formula $\sum_{r = 1}^{n} r^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}$ and simplify

$\begin{array}{rcl} \sum_{r = 1}^{n} {(2 r)}^{3} & = & 8 (\frac{1}{4} n^{2} {(n + 1)}^{2}) \\ = & 2 n^{2} {(n + 1)}^{2} \end{array}$

$\begin{array}{rcl} 2 n^{2} {(n + 1)}^{2} \end{array}$

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Standard Series (Edexcel International A Level Further Maths)

Revision Note

Sums of Natural Numbers

What is sigma notation?

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

What properties of sigma notation do I need to know?

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

Exam Tip

Worked Example

Sums of Squares

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

Exam Tip

Worked Example

Sums of Cubes

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

Exam Tip

Worked Example

You've read 0 of your 0 free revision notes

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Author: Mark Curtis