Sequences and series

Q: What is the factorial of a positive integer \(n\), and what is the special definition for \(0!\)?

The factorial \(n!\) is the product of all positive integers from \(n\) down to \(1\) (i.e. \(n! = n \times (n - 1) \times \dots \times 1\)), and by convention \(0! = 1\).

Q: How is the combinatorial coefficient \(\binom{n}{r}\) defined and evaluated algebraically?

\(\binom{n}{r} = {}^n\text{C}_r = \frac{n!}{r!(n - r)!}\) for integers where \(0 \le r \le n\).

Q: How are the numbers in Pascal's Triangle related to the binomial coefficients \(\binom{n}{r}\)?

The numbers in Row \(n\) of Pascal's Triangle are precisely the values of \(\binom{n}{r}\) for \(r = 0, 1, 2, \dots, n\).

Q: What is the general Binomial Expansion formula for \((a + b)^n\) when \(n\) is a positive integer?

\((a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n\).

Q: What formula gives the general \((r + 1)^{\text{th}}\) term of the expansion \((a + b)^n\)?

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).

Q: Why is it crucial to enclose negative signs and coefficients in brackets when expanding \((a - bx)^n\)?

To ensure powers apply to both the coefficient and the sign correctly, such that \((-bx)^r = (-b)^r x^r\) rather than omitting the sign or power on the constant.

Q: How do you find a specific coefficient in a compound product such as \((p + qx)(a + bx)^n\)?

Expand \((a + bx)^n\) up to the required power, multiply by \((p + qx)\), and collect all cross-terms that produce the desired power of \(x\).

Q: How is a binomial expansion used to approximate a numerical value like \(1.02^5\)?

Equate the binomial base to the numerical base (e.g. \(1 + 2x = 1.02 \implies x = 0.01\)) and substitute that value of \(x\) into the simplified expansion.

Mathematics 2210 · CCEA AS Level · 9 min read · Updated: Aug 31, 2026

1. Introduction to Sequences and Series in AS 1

Welcome to the Sequences and series chapter for CCEA AS 1: Pure Mathematics! While the chapter title mentions sequences and series, under the CCEA AS 1 specification, this unit focuses entirely on one powerful algebraic tool: the Binomial Theorem for positive integer exponents (\(n \in \mathbb{N}\)).

Why is this important? Imagine you need to expand an expression like \((x + 2)^2\). That is easy to do by hand: \((x + 2)(x + 2) = x^2 + 4x + 4\). But what if you need to expand \((x + 2)^8\) or \((2 - 3x)^5\)? Multiplying out eight sets of brackets manually would take forever and lead to messy errors. The Binomial Expansion gives us a fast, reliable shortcut to write down any expanded bracket directly without long multiplication.

Quick Scope Note: In Unit AS 1, we only deal with positive whole-number powers (\(n = 1, 2, 3, \dots\)). Other topics like Arithmetic/Geometric Progressions and negative/fractional powers belong strictly to A2 1.

Key Takeaway: The Binomial Theorem is simply a systematic recipe for multiplying out \((a + b)^n\) quickly and accurately when \(n\) is a positive whole number.

2. The Building Blocks: Factorials, \({}^n\text{C}_r\), and Pascal's Triangle

A. Factorial Notation (\(n!\))

The factorial of a positive integer \(n\), written as \(n!\) (read as "n factorial"), represents the product of all positive integers from \(n\) down to \(1\):
\(n! = n \times (n - 1) \times (n - 2) \times \dots \times 2 \times 1\)

For example:
\(4! = 4 \times 3 \times 2 \times 1 = 24\)
\(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)

Special Definition: By mathematical convention, \(0! = 1\).

B. Combinatorial Coefficients (\({}^n\text{C}_r\) or \(\binom{n}{r}\))

The symbol \(\binom{n}{r}\), also written as \({}^n\text{C}_r\), represents the number of ways to choose \(r\) items from a total of \(n\) distinct items. It is defined as:
\(\binom{n}{r} = {}^n\text{C}_r = \frac{n!}{r!(n - r)!}\) where \(0 \le r \le n\)

Example: To calculate \(\binom{5}{2}\):
\(\binom{5}{2} = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2! \times 3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (3 \times 2 \times 1)} = \frac{20}{2} = 10\)

Helpful Properties of \(\binom{n}{r}\):
1. Boundary values: \(\binom{n}{0} = 1\) and \(\binom{n}{n} = 1\)
2. Symmetry: \(\binom{n}{r} = \binom{n}{n - r}\) (for instance, \(\binom{5}{2} = \binom{5}{3} = 10\))
3. Calculator Tip: You can evaluate \({}^n\text{C}_r\) directly on your scientific calculator using the \(\text{nCr}\) button!

C. Pascal's Triangle

Pascal's Triangle is a visual pyramid of numbers where each entry is the sum of the two numbers directly above it:
Row 0 (\(n = 0\)):              \(1\)
Row 1 (\(n = 1\)):            \(1\)    \(1\)
Row 2 (\(n = 2\)):          \(1\)    \(2\)    \(1\)
Row 3 (\(n = 3\)):        \(1\)    \(3\)    \(3\)    \(1\)
Row 4 (\(n = 4\)):      \(1\)    \(4\)    \(6\)    \(4\)    \(1\)
Row 5 (\(n = 5\)):    \(1\)    \(5\) \(10\) \(10\)    \(5\)    \(1\)

The entries in Row \(n\) are precisely the values of \(\binom{n}{r}\) for \(r = 0, 1, 2, \dots, n\). This links directly to the addition rule: \(\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}\).

Key Takeaway: For small powers like \(n = 3\) or \(n = 4\), Pascal's Triangle provides the coefficients instantly. For larger powers, use the \(\binom{n}{r}\) formula or your calculator.

3. The Binomial Theorem for Positive Integers

A. The General Expansion Formula

For any positive integer \(n \in \mathbb{N}\):
\((a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r\)

Notice the clear pattern across the expansion:
1. The powers of \(a\) start at \(n\) and decrease by \(1\) in each term: \(a^n, a^{n-1}, a^{n-2}, \dots, a^0\).
2. The powers of \(b\) start at \(0\) and increase by \(1\) in each term: \(b^0, b^1, b^2, \dots, b^n\).
3. The sum of the powers of \(a\) and \(b\) in every single term always equals \(n\): \((n - r) + r = n\).
4. An expansion of \((a + b)^n\) always contains exactly \((n + 1)\) terms.

B. The Standard Form \((1 + x)^n\)

When \(a = 1\) and \(b = x\), since \(1\) raised to any power is simply \(1\), the formula simplifies to:
\((1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots + x^n\)

C. The General Term Formula

The \((r + 1)^{\text{th}}\) term of the expansion of \((a + b)^n\) is given by:
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)

Important Note on Term Numbering:
Because the expansion starts with \(r = 0\) (the \(1^{\text{st}}\) term), the value of \(r\) is always one less than the term number:
- For the \(1^{\text{st}}\) term: \(r = 0\)
- For the \(2^{\text{nd}}\) term: \(r = 1\)
- For the \(3^{\text{rd}}\) term: \(r = 2\)
- For the \(4^{\text{th}}\) term: \(r = 3\)

Key Takeaway: \((a + b)^n\) has \(n + 1\) terms. When looking for the \(k^{\text{th}}\) term, always set \(r = k - 1\).

4. Core Exam Question Types & Step-by-Step Examples

Type 1: Direct Expansion of \((a + bx)^n\) in Ascending Powers

Worked Example 1: Expand \((2 - 3x)^4\) fully in ascending powers of \(x\), simplifying all coefficients.

Step 1: Identify \(a\), \(b\), and \(n\).
Here, \(a = 2\), \(b = (-3x)\), and \(n = 4\). Always keep the negative sign attached to the second term inside brackets!

Step 2: Write out the terms using \(\binom{n}{r}\).
\((2 - 3x)^4 = \binom{4}{0}(2)^4(-3x)^0 + \binom{4}{1}(2)^3(-3x)^1 + \binom{4}{2}(2)^2(-3x)^2 + \binom{4}{3}(2)^1(-3x)^3 + \binom{4}{4}(2)^0(-3x)^4\)

Step 3: Evaluate the coefficients and powers carefully.
- Term 1 (\(r = 0\)): \(1 \times 16 \times 1 = 16\)
- Term 2 (\(r = 1\)): \(4 \times 8 \times (-3x) = -96x\)
- Term 3 (\(r = 2\)): \(6 \times 4 \times [(-3)^2 x^2] = 6 \times 4 \times 9x^2 = 216x^2\)
- Term 4 (\(r = 3\)): \(4 \times 2 \times [(-3)^3 x^3] = 4 \times 2 \times (-27x^3) = -216x^3\)
- Term 5 (\(r = 4\)): \(1 \times 1 \times [(-3)^4 x^4] = 1 \times 1 \times 81x^4 = 81x^4\)

Step 4: Combine the simplified terms.
\((2 - 3x)^4 = 16 - 96x + 216x^2 - 216x^3 + 81x^4\)

Type 2: Finding an Unknown Coefficient or Parameter

Worked Example 2: In the expansion of \((1 + kx)^6\), where \(k\) is a positive constant, the coefficient of \(x^2\) is \(60\). Find the value of \(k\).

Step 1: Write down the general term for \(x^2\).
The term involving \(x^2\) occurs when \(r = 2\):
\(T_3 = \binom{6}{2}(1)^{6-2}(kx)^2 = \binom{6}{2}(1)^4 k^2 x^2\)

Step 2: Calculate \(\binom{6}{2}\).
\(\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15\)

Step 3: Express the term and set up the equation.
\(T_3 = 15k^2 x^2\)
The coefficient of \(x^2\) is \(15k^2\). We are given that this coefficient equals \(60\):
\(15k^2 = 60\)

Step 4: Solve for \(k\).
\(k^2 = \frac{60}{15} = 4\)
\(k = \pm \sqrt{4} = \pm 2\)
Since the question states that \(k\) is positive, \(k = 2\).

Type 3: Compound Expansions \((p + qx)(a + bx)^n\)

Worked Example 3: Find the coefficient of \(x^2\) in the expansion of \((3 - x)(1 + 2x)^5\).

Step 1: Expand the binomial part \((1 + 2x)^5\) up to the \(x^2\) term.
\((1 + 2x)^5 = 1 + \binom{5}{1}(2x) + \binom{5}{2}(2x)^2 + \dots\)
\((1 + 2x)^5 = 1 + 5(2x) + 10(4x^2) + \dots\)
\((1 + 2x)^5 = 1 + 10x + 40x^2 + \dots\)

Step 2: Multiply by the linear factor \((3 - x)\).
\((3 - x)(1 + 10x + 40x^2 + \dots)\)

Step 3: Collect only the products that produce an \(x^2\) term.
- Multiplying the constant \(3\) by \(40x^2\): \(3 \times 40x^2 = 120x^2\)
- Multiplying the \(-x\) term by \(10x\): \((-x) \times 10x = -10x^2\)

Step 4: Sum the \(x^2\) terms to get the final coefficient.
\(120x^2 - 10x^2 = 110x^2\)
The coefficient of \(x^2\) is \(110\).

Type 4: Numerical Approximations

Worked Example 4:
(a) Find the first three terms in ascending powers of \(x\) of \((1 + 2x)^5\).
(b) By substituting an appropriate value of \(x\), find an estimate for \(1.02^5\).

Part (a) Solution:
\((1 + 2x)^5 = 1 + \binom{5}{1}(2x) + \binom{5}{2}(2x)^2 + \dots\)
\((1 + 2x)^5 = 1 + 5(2x) + 10(4x^2) + \dots\)
\((1 + 2x)^5 = 1 + 10x + 40x^2 + \dots\)

Part (b) Solution:
We want \((1 + 2x)^5 = 1.02^5\). Set the bases equal:
\(1 + 2x = 1.02 \implies 2x = 0.02 \implies x = 0.01\)

Substitute \(x = 0.01\) into the expansion from part (a):
\(1.02^5 \approx 1 + 10(0.01) + 40(0.01)^2\)
\(1.02^5 \approx 1 + 0.10 + 40(0.0001)\)
\(1.02^5 \approx 1 + 0.10 + 0.004 = 1.104\)

Key Takeaway: For compound products, do not expand everything fully—only track down and combine the specific terms that multiply together to give the requested power of \(x\).

5. Common Pitfalls & Examiner Warnings

Examiners frequently report the same avoidable errors in AS 1 papers. Watch out for these traps:

1. The "Missing Brackets" Error on Coefficients:
When expanding a term like \((3x)^2\), students often write \(3x^2\) instead of \(9x^2\). Always place brackets around both the number and the variable before applying the power: \((bx)^r = b^r x^r\).

2. Losing the Negative Sign:
In an expansion of \((a - bx)^n\), the second term is \((-bx)\). Remember that:
- An odd power of a negative is negative: \((-3x)^3 = -27x^3\)
- An even power of a negative is positive: \((-3x)^2 = +9x^2\)

3. Miscounting the Term Index (\(r\) vs \(r + 1\)):
If a question asks for the \(4^{\text{th}}\) term, do not use \(r = 4\). The expansion starts at \(r = 0\), so the \(4^{\text{th}}\) term uses \(r = 3\).

4. Confusing AS 1 with A2 1 Content:
In AS 1, the power \(n\) is always a positive integer, meaning the expansion has a finite number of terms \((n + 1)\) and terminates. Do not worry about validity ranges like \(|x| < 1\) or infinite series; those belong strictly to Unit A2 1.

5. Forgetting Cross-Terms in Compound Expansions:
When multiplying \((p + qx)(1 + kx)^n\), remember that the \(x^2\) term comes from two sources: the constant times the \(x^2\) term, and the \(x\) term times the \(x\) term. Always collect both!

6. Chapter Summary & Quick Revision Checklist

Use this checklist to ensure you are exam-ready for this chapter:

- Factorials: Can you evaluate \(n!\) and \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\)?
- Pascal's Triangle: Can you write out rows of Pascal's Triangle to find coefficients for small powers?
- Binomial Formula: Do you know \((a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r\)?
- Term Count: Do you remember that \((a + b)^n\) has \(n + 1\) terms?
- General Term: Can you find the \((r+1)^{\text{th}}\) term using \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)?
- Bracket Rule: Did you bracket \((bx)^r = b^r x^r\) to avoid coefficient errors?
- Compound Expansions: Can you find specific coefficients in products like \((a + bx)(1 + kx)^n\)?
- Approximations: Can you choose the correct value of \(x\) to estimate arithmetic powers?

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the factorial of a positive integer \(n\), and what is the special definition for \(0!\)?

The factorial \(n!\) is the product of all positive integers from \(n\) down to \(1\) (i.e. \(n! = n \times (n - 1) \times \dots \times 1\)), and by convention \(0! = 1\).

How is the combinatorial coefficient \(\binom{n}{r}\) defined and evaluated algebraically?

\(\binom{n}{r} = {}^n\text{C}_r = \frac{n!}{r!(n - r)!}\) for integers where \(0 \le r \le n\).

How are the numbers in Pascal's Triangle related to the binomial coefficients \(\binom{n}{r}\)?

The numbers in Row \(n\) of Pascal's Triangle are precisely the values of \(\binom{n}{r}\) for \(r = 0, 1, 2, \dots, n\).

What is the general Binomial Expansion formula for \((a + b)^n\) when \(n\) is a positive integer?

\((a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r}b^r = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n\).

What formula gives the general \((r + 1)^{\text{th}}\) term of the expansion \((a + b)^n\)?

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).

Why is it crucial to enclose negative signs and coefficients in brackets when expanding \((a - bx)^n\)?

To ensure powers apply to both the coefficient and the sign correctly, such that \((-bx)^r = (-b)^r x^r\) rather than omitting the sign or power on the constant.

How do you find a specific coefficient in a compound product such as \((p + qx)(a + bx)^n\)?

Expand \((a + bx)^n\) up to the required power, multiply by \((p + qx)\), and collect all cross-terms that produce the desired power of \(x\).

How is a binomial expansion used to approximate a numerical value like \(1.02^5\)?

Equate the binomial base to the numerical base (e.g. \(1 + 2x = 1.02 \implies x = 0.01\)) and substitute that value of \(x\) into the simplified expansion.

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