Other Sequences (Cambridge (CIE) IGCSE International Maths)

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Amber

Expertise

Maths

Types of Sequences

What are common types of sequences?

Sequences can follow any rule, but common sequences are
- Linear
  - nth term = $a n + b$
- Quadratic
  - nth term = $a n^{2} + b$
- Cubic
  - nth term = $a n^{3} + b$
- Exponential (Geometric)
  - nth term = $a \times r^{n - 1}$
Sequences may also be formed using common numbers
- Prime numbers
  - 2, 3, 5, 7, 11, ...
- Triangular numbers
  - 1, 3, 6, 10, 15, ...

What is a cubic sequence?

A cubic sequence has an n th term formula that involves n³
The third differences are constant (the same)
- These are the differences between the second differences
- For example, 4, 25, 82, 193, 376, 649, ...
  1st Differences: 21, 57, 111, 183, 273, ...
  2nd Differences: 36, 54, 72, 90, ...
  3rd Differences: 18, 18, 18, ...

How do I find the n^th term formula for a cubic sequence?

The sequence with the n th term formula of n³ is the cube numbers
- 1, 8, 27, 64, 125, ...
  - From 1³, 2³, 3³, 4³, ...
Finding the n th term formula of other cubic sequences comes from comparing them to the cube numbers, n³
- 2, 9, 28, 65, 126, ... has the formula n³+ 1
  - Each term is one more than the cube numbers
- 2, 16, 54, 128, 250, ... has the formula 2n³
  - Each term is double a cube number
You can also use second differences to help find the n th term
- For the simple cubic sequence, n th term = an³ + b
  - where a is $\frac{1}{6}$ of the third difference

What is an exponential (geometric) sequence?

An exponential (geometric) sequence is one where you multiply each term by the same number to get the next term
- E.g. 3, 6, 12, 24, 48, ... is exponential because:
  - terms are multiplied by 2 each time
  - 2 is called the common ratio (or constant multiplier)
  - You can find this by dividing any term by the term immediately before, 6 ÷ 3 or 12 ÷ 6 or 24 ÷ 12 etc
The n th term formula is $a \times r^{n - 1}$
- where a is the first term
- r is the common ratio
- n is the position number of the term
  - E.g. 3, 6, 12, 24, 48, ... has a = 3 and r = 2
  - so the n th term = $3 \times 2^{n - 1}$
- Remember that the power is $n - 1$ , not $n$
If the common ratio satisfies
- $r > 1$ then the sequence increases
- $0 < r < 1$ then the sequence decreases
  - E.g. $r = \frac{1}{2}$

How can sequences be made harder?

You may be given a fraction with two different sequences on the top and bottom
- E.g. $\frac{3}{1}, \frac{5}{8}, \frac{7}{27}, \frac{9}{64}, . . .$
  - The numerators are the linear sequence $2 n + 1$
  - The denominators are the cube numbers, $n^{3}$
  - So the n th term formula is $\frac{2 n + 1}{n^{3}}$
You may be asked to find combinations of two different sequences
- E.g. if sequence U is the prime numbers and sequence V has the n th term formula $4 n^{2}$ , find the sequence U + V
  - U = 2, 3, 5, 7, ... and V = 4, 16, 36, 64, ...
  - U + V = (2 + 4), (3 + 16), (5 + 36), (7 + 64), ... = 6, 19, 41, 71, ...
Other problems involving setting up and solving equations
- This may lead to a pair of simultaneous equations

Worked Example

(a) Find the formula for the n^th term of the sequence 5, 19, 57, 131, 253, 435, ...

See if the sequence is linear, quadratic or cubic by finding the first, second or third differences

The first differences are

14, 38, 74, 122, 182

These are not constant, so find the second differences

24, 36, 48, 60

These are not constant, so find the third differences

12, 12, 12

These are constant so the sequence is cubic, with n^th term formula $a n^{3} + b$

Method 1
Use the rule that $a$ is $\frac{1}{6}$ of the third difference

$a = \frac{1}{6} \times 12 = 2$

This means the n^th term formula is $2 n^{3} + b$
There are different ways to find b
E.g. substitute in $n = 1$ to find the first term from the formula, then make it equal to 5 (the first term in the question)

$\begin{array}{rcl} 2 \times 1^{3} + b & = & 5 \\ 2 + b & = & 5 \\ b & = & 3 \end{array}$

The n^th term formula is $2 n^{3} + 3$

Method 2

Compare 5, 19, 57, 131, 253, 435, ... to the cube numbers 1, 8, 27, 64, 125, ...

Double the cube numbers

2, 16, 54, 128, 250, ...

Add 3

5, 19, 57, 131, 253, ...

The cube numbers (with n^th term formula $n^{3}$ ) are doubled then 3 is added

The n^th term formula is $2 n^{3} + 3$

(b) Find the formula for the n^th term of the sequence 4, 20, 100, 500, 2500, ...

Seeing if the first, second or third differences are constant does not work

The numbers are increasing very fast, suggesting it could be exponential
Check to see if each term is multiplied by the same number each time

$4 \times 5 = 20, 20 \times 5 = 100, 100 \times 5 = 500, 500 \times 5 = 2500, . . .$

Each term is multiplied by 5 (the "common ratio") to get the next, so it is exponential
The n^th term formula is $a \times r^{n - 1}$ where $a$ is the first term and $r$ is the common ratio

$a = 4$ and $r = 5$

The n^th term formula is $4 \times 5^{n - 1}$

$\frac{5}{4}, \frac{19}{20}, \frac{57}{100}, \frac{131}{500}, \frac{253}{2500}, . . .$

This sequence is a fraction formed by dividing the sequence in part (a) by the sequence in part (b)
Divide their n^th term formulas

The n^th term formula is $\frac{2 n^{3} + 3}{4 \times 5^{n - 1}}$

Worked Example

The first three terms of an exponential sequence are shown below

$x - 1 2 x x^{2}$

By forming and solving an equation, find the common ratio, $r$ , given that $x \neq 0$ .

As this is an exponential sequence, each term is multiplied by the common ratio, $r$ , to get the next term

Consider the first two terms

$\begin{array}{rcl} (x - 1) \times r & = & 2 x \\ r & = & \frac{2 x}{x - 1} \end{array}$

Considering the next two terms

$\begin{array}{rcl} 2 x \times r & = & x^{2} \\ r & = & \frac{x^{2}}{2 x} \end{array}$

This is a pair of simultaneous equations
They can be solved by substituting one into the other (replacing $r$ )

$\frac{2 x}{x - 1} = \frac{x^{2}}{2 x}$

Multiply both sides by $2 x$

$\frac{4 x^{2}}{x - 1} = x^{2}$

Multiply both sides by $x - 1$ , and expand

$4 x^{2} = x^{2} (x - 1) 4 x^{2} = x^{3} - x^{2}$

Subtract $4 x^{2}$ from both sides and factorise

$0 = x^{3} - 5 x^{2} 0 = x^{2} (x - 5)$

Solve

$\begin{array}{rcl} x^{2} & = & 0 \end{array}$ so $\begin{array}{rcl} x & = & 0 \end{array}$
or
$\begin{array}{rcl} (x - 5) & = & 0 \end{array}$ so $\begin{array}{rcl} x & = & 5 \end{array}$

You are told $x \neq 0$ so $x = 5$
Substitute this into the original sequence, $x - 1, 2 x, x^{2}$

$(5 - 1), 2 (5), 5^{2} = 4, 10, 25$

Find the common ratio $r$ (for example, by dividing a term by its previous term)

$\frac{10}{4} = 2.5$

r = 2.5

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Other Sequences (Cambridge (CIE) IGCSE International Maths)

Revision Note

Types of Sequences

What are common types of sequences?

What is a cubic sequence?

How do I find the nth term formula for a cubic sequence?

What is an exponential (geometric) sequence?

How can sequences be made harder?

Worked Example

Worked Example

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Author: Amber

How do I find the n^th term formula for a cubic sequence?