Sequences: Finding Patterns in the Chaos

Welcome to the chapter on Sequences! This topic is all about spotting patterns in numbers and, crucially, learning how to write a general rule that describes that pattern. Why is this important? Mathematics often uses sequences to model how things grow or change over time, whether it’s bacteria multiplying, money earning interest, or even the arrangement of petals on a flower!

Don't worry if finding these rules seems tricky at first. We will break down the process using simple, reliable methods, especially the 'difference method' which is your best friend here.

1. Understanding the Basics of Sequences

What is a Sequence?

A sequence is simply an ordered list of numbers (or objects) following a specific rule or pattern. Each number in the sequence is called a term.

Example: 2, 4, 6, 8, 10, ...

  • The first term is 2.
  • The fifth term is 10.

We use subscript notation to represent terms. For a sequence $T$:

  • \(T_1\) is the first term.
  • \(T_2\) is the second term.
  • \(T_n\) is the \(n\)th term (the general term/rule).

The Term-to-Term Rule (C2.7.2)

The term-to-term rule tells you how to get from one term to the very next term.

Example 1: 5, 10, 15, 20, ...
The rule is: Add 5 to the previous term.

Example 2: 3, 6, 12, 24, ...
The rule is: Multiply the previous term by 2.

Tip: Continuing a Sequence
To continue a sequence (C2.7.1), just apply the term-to-term rule you found to the last given term.

Quick Review: Special Sequences (C2.7.2)

You must be able to recognise the following common number patterns:

  • Square Numbers: \(n^2\). (1, 4, 9, 16, 25, ...)
  • Cube Numbers: \(n^3\). (1, 8, 27, 64, 125, ...)
  • Triangular Numbers: The pattern of dots needed to form triangles. The rule is \(\frac{n(n+1)}{2}\). (1, 3, 6, 10, 15, ...)

2. Linear Sequences (Arithmetic Progressions)

A sequence is linear (or arithmetic) if the difference between consecutive terms is always the same. This constant difference is called the common difference, \(d\).

Finding the \(n\)th Term Rule (\(T_n\)) (C2.7.3(a))

The general formula for the \(n\)th term of a linear sequence is:

$$T_n = an + b$$

Where \(a\) is the common difference, and \(b\) is related to the 'zeroth' term (the term before \(T_1\)).

Step-by-Step Method:

Example: Find the \(n\)th term of 4, 7, 10, 13, ...

  1. Find the Common Difference (\(a\)):
    • \(7 - 4 = 3\)
    • \(10 - 7 = 3\)

    The common difference \(a = 3\). So, the rule starts with \(T_n = 3n\).

  2. Compare to the \(3n\) sequence:

    Write out the sequence \(3n\): 3, 6, 9, 12, ...

  3. Find the Adjustment (\(b\)):

    Compare the original sequence (4, 7, 10, 13, ...) to the \(3n\) sequence (3, 6, 9, 12, ...). You must add 1 to every term.

    (e.g., \(4 = 3 + 1\); \(7 = 6 + 1\))

    The adjustment \(b = 1\).

  4. Write the Rule:

    $$T_n = 3n + 1$$

Memory Aid: For Linear Sequences, the coefficient of n (which is a) is always the first difference.

Key Takeaway for Linear Sequences

Linear sequences have a constant first difference. The \(n\)th term formula \(T_n = an + b\) is easy to find by calculating the difference (\(a\)) and finding the necessary adjustment (\(b\)).

3. Quadratic Sequences

A sequence is quadratic when the second difference (the difference of the differences) is constant. Quadratic sequences are defined by the general formula:

$$T_n = an^2 + bn + c$$

Finding the \(n\)th Term for Quadratic Sequences (C2.7.3(b))

This relies on the powerful Difference Method (C2.7 syllabus explicitly mentions using this method).

Step-by-Step Method (Difference Method):

Example: Find the \(n\)th term of 2, 5, 10, 17, ...

  1. Find the First Differences:

    Differences between consecutive terms: 3, 5, 7, ... (Not constant, so not linear.)

  2. Find the Second Differences:

    Differences between the first differences: 2, 2, ... (This is constant! So it is quadratic.)

  3. Determine \(a\) (the coefficient of \(n^2\)):

    The constant second difference is equal to \(2a\).

    $$2a = 2$$ $$a = 1$$

    So, the rule starts with \(T_n = 1n^2 + bn + c\).

  4. Determine \(b\) and \(c\) (by creating a new sequence):

    Subtract the \(n^2\) sequence (1, 4, 9, 16, ...) from the original sequence.

    Original Sequence (\(T_n\)) 2 5 10 17
    \(n^2\) Sequence 1 4 9 16
    Remaining Sequence (\(T_n - n^2\)) 1 1 1 1

    The remaining sequence is 1, 1, 1, 1, ...

  5. Find the \(n\)th term for the Remaining Sequence:

    This remaining sequence is a simple linear sequence: \(1n^0 + 1\) or just \(1\). In the form \(bn + c\), since the difference is 0, \(b=0\), and \(c=1\).

    So, the linear part is \(0n + 1\).

  6. Combine the Parts:

    The full \(n\)th term is the \(n^2\) part plus the linear part:

    $$T_n = n^2 + 1$$

Common Mistake to Avoid: Always remember to halve the second difference to find \(a\). If the second difference is 6, then \(a=3\).

Key Takeaway for Quadratic Sequences

Quadratic sequences have a constant second difference. Use the relationship \(2a = \text{Second Difference}\) to find the \(n^2\) term, and then subtract this pattern to find the remaining linear part \(bn + c\).

4. Cubic Sequences

A sequence is cubic (E2.7.3/C2.7.3(c) for simple cubic) when the third difference is constant. Cubic sequences are defined by the general formula:

$$T_n = an^3 + bn^2 + cn + d$$

Finding the \(n\)th Term for Cubic Sequences (E2.7.3)

The difference method extends here, involving three levels of differences:

  • First difference: \(T_2 - T_1\)
  • Second difference: Difference of the first differences
  • Third difference: Difference of the second differences (This is constant!)
The Core Relationships:

If the third difference is constant, we use the following relationships to find the coefficients:

  • Third Difference \(= 6a\)
  • First term of the Second Difference row \(= 12a + 2b\) (This step is often skipped in IGCSE using the simpler subtraction method below)
  • First term of the First Difference row \(= 7a + 3b + c\)
  • First Term \(T_1\) \(= a + b + c + d\)

Easier Method (Subtraction):
For IGCSE, especially for "simple cubic" (C2.7) or general cubic (E2.7), the easiest method is similar to the quadratic technique:

  1. Find the constant third difference and calculate \(a\) using \(6a = \text{Third Difference}\).
  2. Generate the \(an^3\) sequence.
  3. Subtract the \(an^3\) sequence from the original sequence.
  4. The remaining sequence will be quadratic (\(bn^2 + cn + d\)).
  5. Use the Quadratic Difference Method (Section 3) on the remaining sequence to find \(b, c,\) and \(d\).

Did you know?
This method of taking differences works because if you subtract a cubic pattern (\(an^3\)) from a cubic sequence, you are left with a simpler sequence (a quadratic one).

Key Takeaway for Cubic Sequences

Cubic sequences have a constant third difference. Use \(6a = \text{Third Difference}\), then subtract the \(an^3\) pattern to simplify the remaining sequence into a quadratic one.

5. Exponential Sequences (Extended Content E2.7 only)

Exponential sequences are related to multiplication, not addition. They have a constant ratio between consecutive terms, called the common ratio, \(r\).

This type of sequence is also known as a Geometric Progression.

Example: 3, 6, 12, 24, ...
Here, \(r = 2\) (you multiply by 2 each time).

Finding the \(n\)th Term for Exponential Sequences (E2.7)

The general formula for the \(n\)th term of an exponential sequence is:

$$T_n = ar^{n-1}$$

Where \(a\) is the first term, and \(r\) is the common ratio.

Step-by-Step Method:

Example: Find the \(n\)th term of 5, 10, 20, 40, ...

  1. Identify the First Term (\(a\)):

    $$a = 5$$

  2. Find the Common Ratio (\(r\)):

    Divide any term by the previous term: \(10/5 = 2\), \(20/10 = 2\).

    $$r = 2$$

  3. Write the Rule:

    Substitute \(a=5\) and \(r=2\) into the formula \(T_n = ar^{n-1}\):

    $$T_n = 5(2)^{n-1}$$

Analogy: Exponential growth is like viral spreading or compound interest. The amount of increase gets larger every time, because you are multiplying by a ratio, not just adding a constant value.

Key Takeaway for Exponential Sequences (Extended)

Exponential sequences have a constant ratio (\(r\)). The \(n\)th term uses the formula \(T_n = ar^{n-1}\), where \(a\) is the first term.

6. Summary and Exam Strategies

Checking Your Sequence Type

When you encounter a new sequence, use this hierarchy to find the rule:

  1. Check Ratios: Is there a constant ratio? (If yes, it’s Exponential/Geometric - E2.7)
  2. Check First Difference: Is the first difference constant? (If yes, it’s Linear/Arithmetic: \(T_n = an + b\))
  3. Check Second Difference: Is the second difference constant? (If yes, it’s Quadratic: \(T_n = an^2 + bn + c\))
  4. Check Third Difference: Is the third difference constant? (If yes, it’s Cubic: \(T_n = an^3 + ...\))

Using the \(n\)th Term Rule

Once you have the \(n\)th term rule, you can find any term in the sequence without listing all the preceding numbers.

Example: If \(T_n = n^2 + 1\), find the 20th term.
Substitute \(n=20\): \(T_{20} = (20)^2 + 1 = 400 + 1 = 401\).

🌟 Final Study Tip 🌟

Always practice the 'difference method' rigorously. It is the core mathematical technique required for solving linear, quadratic, and cubic \(n\)th term problems in IGCSE International Mathematics (0607). Remember the key starting relationships:

  • Linear: \(a = \text{First Difference}\)
  • Quadratic: \(2a = \text{Second Difference}\)
  • Cubic: \(6a = \text{Third Difference}\)