Sequences: Finding Order in Numerical Patterns
Hello mathematicians! This chapter is all about sequences—ordered lists of numbers that follow a specific rule or pattern. Think of sequences as secret codes where you need to find the rule to predict the next numbers.
Sequences are a fundamental part of the "Algebra and graphs" section because the rules that govern them are usually expressed as algebraic formulas (the \(n^{th}\) term). Mastering this topic will help you predict future values and describe complex patterns efficiently.
✦ Section 1: Introduction to Sequences and Term-to-Term Rules
1.1 Key Vocabulary
A sequence is simply an ordered list of numbers.
- The numbers in the list are called terms.
- We usually label the position of a term using the letter \(n\), where \(n\) is always a positive integer (\(1, 2, 3, \dots\)).
Example Sequence: 2, 4, 6, 8, 10, ...
Here, the 1st term is 2, the 2nd term is 4, and so on.
1.2 The Term-to-Term Rule
The term-to-term rule tells you how to get from one term to the next.
- Example: In the sequence 5, 8, 11, 14, ... the rule is "Add 3".
- Example: In the sequence 40, 20, 10, 5, ... the rule is "Divide by 2" (or multiply by 0.5).
Important Note: While the term-to-term rule is great for finding the next number, the \(n^{th}\) term rule is much more powerful because it lets you jump straight to the 100th term without calculating all the terms before it!
1.3 Recognising Special Sequences (Patterns)
You must be able to recognise and continue sequences related to common numbers:
Square Numbers (\(n^2\))
1, 4, 9, 16, 25, ... (These are numbers formed by squaring \(n\)).
Cube Numbers (\(n^3\))
1, 8, 27, 64, 125, ... (These are numbers formed by cubing \(n\)).
Triangular Numbers
1, 3, 6, 10, 15, 21, ... (These represent the number of dots needed to form a triangle. The difference between consecutive terms increases by 1 each time).
Did you know? The \(n^{th}\) term for triangular numbers is \(\frac{n(n+1)}{2}\).
Key Takeaway: Sequences are ordered lists. The term-to-term rule links adjacent terms, but the \(n^{th}\) term rule (algebraic) is needed to find any term quickly.
✦ Section 2: Linear Sequences (Finding the \(n^{th}\) Term)
2.1 What is a Linear Sequence?
A linear sequence (also called an Arithmetic Progression) is a sequence where the difference between consecutive terms is constant. This constant difference is called the common difference, \(d\).
The algebraic form (the \(n^{th}\) term) for any linear sequence is always:
\[T_n = an + b\]
where \(a\) is the common difference, and \(b\) is the imaginary "zero term" (the term before the first term).
2.2 Step-by-Step: Finding the \(n^{th}\) Term for Linear Sequences
Let's find the \(n^{th}\) term for the sequence: 7, 10, 13, 16, ...
- Find the Common Difference (\(a\)):
The difference between each term is constant: \(10-7=3\), \(13-10=3\), etc.
So, \(a = 3\). The formula starts with \(T_n = 3n\). - Find the Zero Term (\(b\)):
Look at your new formula, \(3n\). Compare the terms in the sequence to the results of \(3n\):- For \(n=1\), \(3n = 3(1) = 3\). But the actual term is 7. (Difference is \(7-3 = 4\))
- For \(n=2\), \(3n = 3(2) = 6\). But the actual term is 10. (Difference is \(10-6 = 4\))
(Alternatively, work backwards: if the first term is 7 and you add 3 each time, the term before 7 must be \(7 - 3 = 4\)). - Write the Final Formula:
\[T_n = 3n + 4\]
Quick Check: Test the formula for the 4th term (\(n=4\)): \(3(4) + 4 = 12 + 4 = 16\). Correct!
👉 Common Mistake Alert!
Students often forget to find the zero term (\(b\)). If the common difference is 5, you write \(5n\) and stop. Remember: \(5n\) generates the sequence 5, 10, 15, ... If your sequence is 8, 13, 18, ..., you must add 3 to shift the entire sequence up!
Key Takeaway: The \(n^{th}\) term of a linear sequence is \(T_n = an + b\), where \(a\) is the common difference, and \(b\) is the "zero term".
✦ Section 3: Quadratic Sequences
3.1 Recognising a Quadratic Sequence
A sequence is quadratic if the first difference is not constant, but the second difference (the difference between the differences) is constant.
The algebraic form for a quadratic sequence is:
\[T_n = an^2 + bn + c\]
3.2 Step-by-Step: Finding the \(n^{th}\) Term for Quadratic Sequences
This method (often called the Difference Method) involves a set of rules related to the coefficients \(a, b, \text{ and } c\).
Let's find the \(n^{th}\) term for the sequence: 3, 7, 13, 21, 31, ...
- Calculate the First and Second Differences:
Sequence (\(T_n\)): 3, 7, 13, 21, 31
1st Difference: 4, 6, 8, 10
2nd Difference: 2, 2, 2 - Find Coefficient \(a\):
The constant second difference is equal to \(2a\).
\[2a = 2 \implies a = 1\]
So, the formula starts with \(T_n = 1n^2 \dots\) - Find Coefficient \(b\):
The first difference between the 1st and 2nd terms (the first number in your 1st Difference row, which is 4) is related to \(a\) and \(b\) by the formula \(3a + b\).
\[3a + b = \text{First term of 1st Difference}\]
Since \(a=1\):
\[3(1) + b = 4 \implies 3 + b = 4 \implies b = 1\]
So, the formula is now \(T_n = 1n^2 + 1n \dots\) - Find Coefficient \(c\):
The first term of the sequence (3) is related to \(a, b,\) and \(c\) by the formula \(a + b + c\).
\[a + b + c = \text{First Term of Sequence}\]
Since \(a=1\) and \(b=1\):
\[1 + 1 + c = 3 \implies 2 + c = 3 \implies c = 1\] - Write the Final Formula:
\[T_n = n^2 + n + 1\]
Quick Review Box (Quadratic):
1. 2nd Difference \( = 2a\)
2. First term of 1st Difference \( = 3a + b\)
3. First Term of Sequence \( = a + b + c\)
Key Takeaway: If the second difference is constant, the sequence is quadratic. You must use the three algebraic rules (\(2a\), \(3a+b\), \(a+b+c\)) to find the coefficients \(a, b,\) and \(c\).
✦ Section 4: Simple Cubic Sequences (Extended Syllabus)
(This section is required for Extended candidates.)
4.1 Recognising a Cubic Sequence
A sequence is cubic if the constant difference is found in the third difference row.
The algebraic form for a cubic sequence is:
\[T_n = an^3 + bn^2 + cn + d\]
4.2 Step-by-Step: Finding the \(n^{th}\) Term for Simple Cubic Sequences
The difference method extends naturally to cubic sequences.
Rules for Cubic Sequences:
- 3rd Difference \( = 6a\)
- First term of 2nd Difference \( = 12a + 2b\) (or \(6a + 2b\))
- First term of 1st Difference \( = 7a + 3b + c\)
- First Term of Sequence \( = a + b + c + d\)
Don't worry if those formulas look scary! You can often solve "simple cubic" problems by using a slightly easier method: comparing to the base cubic sequence.
Alternative Method (Comparison)
Let's find the \(n^{th}\) term for the sequence: 5, 12, 31, 68, 133, ...
- Check the differences: (7, 19, 37, 65). 2nd Diff: (12, 18, 28). 3rd Diff: (6, 6).
Since the 3rd difference is 6, we know \(6a=6\), so \(a=1\). The sequence is approximately \(n^3\). - Compare to \(n^3\):
\(n\): 1, 2, 3, 4, 5
\(T_n\): 5, 12, 31, 68, 133
\(n^3\): 1, 8, 27, 64, 125
Difference: 4, 4, 4, 4, 8 (Wait, something is wrong with the example data, let's use a cleaner one!)Revised Example: Sequence: 2, 9, 28, 65, 126, ...
\(n\): 1, 2, 3, 4, 5
\(T_n\): 2, 9, 28, 65, 126
\(n^3\): 1, 8, 27, 64, 125
Difference (Remainder Sequence): 1, 1, 1, 1, 1 - Find the \(n^{th}\) term of the Remainder Sequence:
The remainder sequence is 1, 1, 1, 1, ... This is simply the constant 1. - Combine the Terms:
\[T_n = n^3 + 1\]
Encouragement: For IGCSE, simple cubic sequences often follow this structure where the remainder is a simple linear sequence or a constant. Look for these patterns first!
Key Takeaway: Cubic sequences have a constant third difference. For simple examples, try subtracting the known \(an^3\) component (usually \(n^3\)) and finding the rule for the remaining sequence.
✦ Section 5: Exponential Sequences and Combinations (Extended Syllabus)
(This section is required for Extended candidates.)
5.1 Exponential Sequences
An exponential sequence (also known as a Geometric Progression) is a sequence where you multiply by a constant amount to get to the next term. This constant multiplier is called the common ratio, \(r\).
The algebraic form for an exponential sequence is:
\[T_n = ar^{n-1}\]
(where \(a\) is the first term, and \(r\) is the common ratio.)
Example: 3, 6, 12, 24, 48, ...
- Common Ratio, \(r\): 2 (you multiply by 2 each time)
- First term, \(a\): 3
- \(n^{th}\) term: \(T_n = 3(2^{n-1})\)
Checking: \(T_3 = 3(2^{3-1}) = 3(2^2) = 3 \times 4 = 12\). Correct!
5.2 Sequences as Combinations of Rules
Sometimes a sequence is a combination of two basic rules (e.g., quadratic plus a linear term, or exponential plus a constant).
Example: Find the \(n^{th}\) term of the sequence: 4, 7, 12, 19, 28, ...
If you check the differences, you find the 2nd difference is 2. This means the sequence is quadratic and \(2a=2\), so \(a=1\). The sequence starts with \(n^2\).
Comparison Method for Combinations:
Sequence (\(T_n\)): 4, 7, 12, 19, 28
Known part (\(n^2\)): 1, 4, 9, 16, 25
Remainder: 3, 3, 3, 3, 3
The remainder sequence is the constant 3. Therefore, the combined \(n^{th}\) term is:
\[T_n = n^2 + 3\]
This skill requires good pattern recognition. If the difference method gets complicated, think about what basic pattern (\(n^2\), \(n^3\), or \(2^n\)) might be embedded in the sequence.
Key Takeaway: Exponential sequences use multiplication (a common ratio). Complex sequences can often be broken down into a sum of two simpler sequences (like quadratic + constant).
✦ Summary of \(n^{th}\) Term Rules
Algebraic rules are your toolkit for sequences. Use this quick reference:
| Sequence Type | Difference Level | General Form of \(T_n\) |
|---|---|---|
| Linear (Arithmetic) | 1st difference is constant (a) | \(an + b\) |
| Quadratic | 2nd difference is constant (2a) | \(an^2 + bn + c\) |
| Cubic | 3rd difference is constant (6a) | \(an^3 + bn^2 + cn + d\) |
| Exponential (Geometric) | Ratio is constant (r) | \(ar^{n-1}\) |
By understanding the difference method, you hold the key to solving nearly all sequence problems in this curriculum!