Sequences and Series: The Binomial Expansion (AS 1)
Welcome to the Sequences and Series chapter of Pure Mathematics AS 1! In this unit, our primary focus is mastering the Binomial Expansion for positive integer powers.
Have you ever tried multiplying out \((x + y)^2\)? That is quite easy: \(x^2 + 2xy + y^2\). But what about \((x + y)^8\) or \((2 - 3x)^5\)? Multiplying brackets by hand would take forever and it is very easy to make mistakes. Fortunately, mathematicians discovered an elegant pattern that allows us to jump straight to the answer. In this chapter, you will learn the rules and tricks to expand any bracket of the form \((a + b)^n\) quickly and accurately.
Don't worry if algebra sometimes feels overwhelming! We will break everything down into simple, step-by-step stages so you can tackle these exam questions with confidence.
1. The Building Blocks: Factorials and Combinations
What is a Factorial?
The factorial of a positive whole number \(n\), written as \(n!\) (read as "n factorial"), is simply the product of all positive integers from \(n\) down to \(1\).
\(n! = n \times (n - 1) \times (n - 2) \times \dots \times 3 \times 2 \times 1\)
Examples:
• \(3! = 3 \times 2 \times 1 = 6\)
• \(4! = 4 \times 3 \times 2 \times 1 = 24\)
• \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\)
Important Special Case: By mathematical definition, \(0! = 1\). This might look strange, but it ensures that our formulas work consistently every time!
What is a Binomial Coefficient?
A binomial coefficient is written as \(\binom{n}{r}\) or \({^n}C_r\) (often read as "n choose r"). It represents the number of ways to choose \(r\) items from a total set of \(n\) items.
The mathematical formula is:
\(\binom{n}{r} = \frac{n!}{r!(n - r)!}\)
Example: Calculate \(\binom{5}{2}\).
\(\binom{5}{2} = \frac{5!}{2!(5 - 2)!} = \frac{5!}{2! \times 3!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{(2 \times 1) \times (3 \times 2 \times 1)} = \frac{20}{2} = 10\)
Calculator Tip: On your scientific calculator, look for the \(\text{nCr}\) button (often found above the division key \(\div\)). To compute \(\binom{5}{2}\), type \(5\), then press \(\text{nCr}\), then \(2\), and press \(=\).
Pascal's Triangle
Before calculators existed, Blaise Pascal popularized a triangular arrangement of numbers where each number is the sum of the two numbers directly above it:
Row \(0\): \(1\)
Row \(1\): \(1\) \(1\)
Row \(2\): \(1\) \(2\) \(1\)
Row \(3\): \(1\) \(3\) \(3\) \(1\)
Row \(4\): \(1\) \(4\) \(6\) \(4\) \(1\)
Row \(5\): \(1\) \(5\) \(10\) \(10\) \(5\) \(1\)
Did you know? The numbers in Row \(n\) are exactly the values of \(\binom{n}{r}\) for \(r = 0, 1, 2, \dots, n\). For instance, Row \(4\) gives \(\binom{4}{0}=1\), \(\binom{4}{1}=4\), \(\binom{4}{2}=6\), \(\binom{4}{3}=4\), \(\binom{4}{4}=1\).
Symmetry Property: Notice that the rows are symmetrical. That means:
\(\binom{n}{r} = \binom{n}{n - r}\)
For example, \(\binom{7}{2} = \binom{7}{5} = 21\).
Section 1 Quick Review
• \(n!\) multiplies all integers from \(n\) down to \(1\).
• \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\) gives the coefficients in the expansion.
• Always remember that \(0! = 1\) and \(\binom{n}{0} = \binom{n}{n} = 1\).
2. The Binomial Expansion for Positive Integer \(n\)
The word binomial simply means "two terms" (like a bicycle has two wheels). A binomial expression looks like \((a + b)\).
The General Formula
For any positive integer \(n\):
\((a + b)^n = \binom{n}{0}a^n + \binom{n}{1}a^{n-1}b^1 + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{r}a^{n-r}b^r + \dots + \binom{n}{n}b^n\)
Since \(\binom{n}{0} = 1\) and \(\binom{n}{1} = n\), we can also write it as:
\((a + b)^n = a^n + n a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n\)
Key Patterns to Notice
1. The power of the first term (\(a\)) starts at \(n\) and decreases by \(1\) in each term until it reaches \(0\).
2. The power of the second term (\(b\)) starts at \(0\) and increases by \(1\) in each term until it reaches \(n\).
3. The sum of the powers in every term always equals \(n\). For instance, in \(a^{n-r}b^r\), \((n - r) + r = n\).
4. An expansion of power \(n\) always contains exactly \((n + 1)\) terms.
Step-by-Step Example 1: Basic Expansion
Problem: Expand \((x + 2)^4\) in ascending powers of \(x\).
Solution:
Here, \(a = x\), \(b = 2\), and \(n = 4\).
Using the formula:
\((x + 2)^4 = \binom{4}{0}x^4(2)^0 + \binom{4}{1}x^3(2)^1 + \binom{4}{2}x^2(2)^2 + \binom{4}{3}x^1(2)^3 + \binom{4}{4}x^0(2)^4\)
Now evaluate each component:
• First term: \(1 \times x^4 \times 1 = x^4\)
• Second term: \(4 \times x^3 \times 2 = 8x^3\)
• Third term: \(6 \times x^2 \times 4 = 24x^2\)
• Fourth term: \(4 \times x \times 8 = 32x\)
• Fifth term: \(1 \times 1 \times 16 = 16\)
Putting it together:
\((x + 2)^4 = x^4 + 8x^3 + 24x^2 + 32x + 16\)
Written in ascending powers of \(x\): \(16 + 32x + 24x^2 + 8x^3 + x^4\).
Handling Negative Signs and Coefficients
Warning: A very common exam mistake is forgetting to place brackets around negative terms or coefficients!
If the expression is \((2 - 3x)^4\), treat \(a = 2\) and \(b = (-3x)\). When raised to a power, \((-3x)^2 = +9x^2\) and \((-3x)^3 = -27x^3\).
Step-by-Step Example 2: Negative and Scaled Terms
Problem: Find the first four terms in the expansion of \((1 - 2x)^5\) in ascending powers of \(x\).
Solution:
Here, \(a = 1\), \(b = (-2x)\), and \(n = 5\).
\((1 - 2x)^5 = \binom{5}{0}(1)^5 + \binom{5}{1}(1)^4(-2x)^1 + \binom{5}{2}(1)^3(-2x)^2 + \binom{5}{3}(1)^2(-2x)^3 + \dots\)
Evaluate term-by-term:
• Term 1: \(1 \times 1 = 1\)
• Term 2: \(5 \times 1 \times (-2x) = -10x\)
• Term 3: \(10 \times 1 \times (4x^2) = +40x^2\)
• Term 4: \(10 \times 1 \times (-8x^3) = -80x^3\)
Therefore, the first four terms are:
\((1 - 2x)^5 = 1 - 10x + 40x^2 - 80x^3 + \dots\)
Section 2 Quick Review
• Use \((a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r\).
• Always use brackets around negative terms like \((-2x)^r\).
• Double-check whether the question asks for ascending powers (smallest power first) or descending powers (largest power first).
3. Finding Specific Terms and Unknown Constants
In CCEA AS 1 exam questions, you are often asked to find a single term (such as "the term independent of \(x\)" or "the coefficient of \(x^3\)") without expanding the entire bracket.
The General Term Formula
The \((r + 1)^{\text{th}}\) term of the expansion of \((a + b)^n\) is given by:
\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
Why \((r + 1)\)?
Because when \(r = 0\), we get the 1st term: \(T_1 = \binom{n}{0}a^n b^0\).
When \(r = 1\), we get the 2nd term: \(T_2 = \binom{n}{1}a^{n-1}b^1\), and so on.
Example 3: Finding a Specific Coefficient
Problem: Find the coefficient of \(x^3\) in the expansion of \((3 + 2x)^7\).
Solution:
Step 1: Identify \(a = 3\), \(b = 2x\), and \(n = 7\).
Step 2: Write down the general term: \(\binom{7}{r} (3)^{7-r} (2x)^r\).
Step 3: To get \(x^3\), we need \(r = 3\).
Step 4: Substitute \(r = 3\):
\(T_4 = \binom{7}{3} \times 3^{7-3} \times (2x)^3\)
\(\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35\)
\(3^4 = 81\)
\((2x)^3 = 8x^3\)
\(T_4 = 35 \times 81 \times 8x^3 = 22680x^3\)
Answer: The coefficient of \(x^3\) is \(22680\). (Note: Write just the number if asked for the coefficient, not the \(x^3\) part!)
Example 4: Working Backwards to Find an Unknown Constant
Problem: In the binomial expansion of \((2 + kx)^6\), the coefficient of \(x^2\) is \(2160\). Given that \(k > 0\), find the value of the constant \(k\).
Solution:
Step 1: Identify the general term: \(\binom{6}{r} (2)^{6-r} (kx)^r\).
Step 2: For the \(x^2\) term, set \(r = 2\):
\(\text{Term} = \binom{6}{2} \times 2^4 \times (kx)^2\)
Step 3: Calculate the numerical parts:
\(\binom{6}{2} = 15\)
\(2^4 = 16\)
\(\text{Term} = 15 \times 16 \times k^2 x^2 = 240 k^2 x^2\)
Step 4: Set the coefficient equal to \(2160\):
\(240 k^2 = 2160\)
\(k^2 = \frac{2160}{240} = 9\)
\(k = \pm 3\)
Since the question states \(k > 0\), we have \(k = 3\).
Section 3 Quick Review
• The general term is \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\).
• For an \(x^r\) term, make sure the power applied to the \(x\)-component matches \(r\).
• "Coefficient" means the numerical factor in front of the \(x\) power.
4. Approximations Using the Binomial Expansion
One powerful practical application of the Binomial Expansion is approximating complicated powers of decimals without using a long calculation.
Key Concept: If \(x\) is a very small number (for example, \(x = 0.01\)), then:
• \(x^2 = 0.0001\)
• \(x^3 = 0.000001\)
Because higher powers of small numbers become tiny very quickly, the first few terms of the expansion give an extremely accurate approximation!
Example 5: Approximating a Decimal Power
Problem:
(a) Expand \((1 + x)^5\) up to and including the term in \(x^3\).
(b) By substituting a suitable value for \(x\), find an estimate for \((1.02)^5\).
Solution:
Part (a):
\((1 + x)^5 = 1 + \binom{5}{1}x + \binom{5}{2}x^2 + \binom{5}{3}x^3 + \dots\)
\((1 + x)^5 = 1 + 5x + 10x^2 + 10x^3 + \dots\)
Part (b):
We want \((1 + x) = 1.02\), which means \(x = 0.02\).
Substitute \(x = 0.02\) into our expansion from part (a):
\((1.02)^5 \approx 1 + 5(0.02) + 10(0.02)^2 + 10(0.02)^3\)
• \(5(0.02) = 0.1\)
• \(10(0.0004) = 0.004\)
• \(10(0.000008) = 0.00008\)
Add them all together:
\((1.02)^5 \approx 1 + 0.1 + 0.004 + 0.00008 = 1.10408\)
Checking with a calculator: \((1.02)^5 = 1.1040808\dots\) Our approximation is correct to \(5\) decimal places using just four terms!
5. Common Mistakes to Avoid in CCEA AS 1 Exams
1. Dropping Brackets with Powers:
Writing \(2x^3\) instead of \((2x)^3 = 8x^3\) is the single most common error. Always write parentheses around terms before evaluating.
2. Mishandling Minus Signs:
Remember: \((-3x)^2 = +9x^2\), but \((-3x)^3 = -27x^3\). An even power gives a positive value; an odd power preserves the negative sign.
3. Confusing Term Number and \(r\):
The 3rd term has \(r = 2\), not \(r = 3\). Always remember that the index \(r\) starts at \(0\).
4. Including \(x\) When Asked for the Coefficient:
If a question asks for the "coefficient of \(x^2\)", the answer is a number (e.g. \(40\)), not \(40x^2\).
Chapter Summary & Key Formulas
• Factorials: \(n! = n \times (n - 1) \times \dots \times 1\), with \(0! = 1\).
• Binomial Coefficient: \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\)
• Binomial Theorem: \((a + b)^n = \sum_{r=0}^n \binom{n}{r} a^{n-r} b^r\)
• Standard \((1 + x)^n\) Form: \((1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dots + x^n\)
• General Term: \(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)
Well done! You have covered all the key concepts for Sequences and Series in AS 1 Pure Mathematics. Practice several past paper questions to build your speed and accuracy!