Mathematics Form 4 | Binomial Expansion | Masomotele Free Learning

📋 Learning Outcomes

By the end of this lesson, you should be able to:

Expand binomial expressions using Pascal's Triangle
State and apply the Binomial Theorem
Find specific terms in a binomial expansion
Use binomial expansion to make approximations
Apply binomial expansion to real-life problems
Identify patterns in binomial coefficients
Use nCr notation to find coefficients

🔍 Key Inquiry Question: How can we efficiently expand expressions like (1 + x)ⁿ without multiplying repeatedly?

🔺 Pascal's Triangle

Building Pascal's Triangle

Each number is the sum of the two numbers above it. The rows give the coefficients of binomial expansions.

1

11

121

1331

14641

15101051

Row n: gives coefficients for (a+b)ⁿ. Row 0 → (a+b)⁰, Row 4 → (a+b)⁴ coefficients are 1,4,6,4,1

Using Pascal's Triangle

Expand (a + b)⁴:

Coefficients from row 4: 1, 4, 6, 4, 1

(a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

📐 Binomial Theorem

The Theorem

(a + b)ⁿ = Σ ⁿCᵣ · aⁿ⁻ʳ · bʳ (r = 0 to n)

ⁿCᵣ = n! / (r!(n−r)!)

nCr (read "n choose r") is the binomial coefficient — the number of ways to choose r items from n.

Computing nCr

Expression	Calculation	Value
⁵C₀	5!/(0!5!)	1
⁵C₂	5!/(2!3!) = (5×4)/(2×1)	10
⁵C₃	5!/(3!2!)	10
⁶C₄	6!/(4!2!) = (6×5)/(2×1)	15

⚠️ Note: ⁿCᵣ = ⁿCₙ₋ᵣ (symmetry property)

✏️ Expanding Using Binomial Theorem

Worked Examples

Example 1: Expand (1 + x)⁵

(1+x)⁵ = ⁵C₀(1)⁵ + ⁵C₁(1)⁴x + ⁵C₂(1)³x² + ⁵C₃(1)²x³ + ⁵C₄(1)x⁴ + ⁵C₅x⁵

= 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵

Example 2: Expand (2 + 3x)⁴

a=2, b=3x, n=4

= ⁴C₀(2)⁴ + ⁴C₁(2)³(3x) + ⁴C₂(2)²(3x)² + ⁴C₃(2)(3x)³ + ⁴C₄(3x)⁴

= 16 + 4(8)(3x) + 6(4)(9x²) + 4(2)(27x³) + 81x⁴

= 16 + 96x + 216x² + 216x³ + 81x⁴

Example 3: Expand (x − 2)³

a=x, b=−2, n=3

= x³ + 3x²(−2) + 3x(−2)² + (−2)³

= x³ − 6x² + 12x − 8

🎯 Finding a Specific Term

General Term Formula

Tᵣ₊₁ = ⁿCᵣ · aⁿ⁻ʳ · bʳ

The (r+1)th term in the expansion of (a+b)ⁿ

Example 1: Find the 4th term of (2+x)⁶

4th term → r = 3

T₄ = ⁶C₃ · (2)³ · x³ = 20 · 8 · x³ = 160x³

Example 2: Find the term containing x⁴ in (1+2x)⁷

Tᵣ₊₁ = ⁷Cᵣ(1)⁷⁻ʳ(2x)ʳ = ⁷Cᵣ · 2ʳ · xʳ

For x⁴: r = 4

T₅ = ⁷C₄ · 2⁴ · x⁴ = 35 · 16 · x⁴ = 560x⁴

🔢 Approximations Using Binomial

Using Binomial for Approximations

When x is small, higher powers of x are negligible. We can use the first few terms to approximate.

Example 1: Approximate (1.02)⁶

Write as (1 + 0.02)⁶

≈ 1 + 6(0.02) + 15(0.02)² + 20(0.02)³ ...

≈ 1 + 0.12 + 0.006 + 0.00016

≈ 1.1262 (actual: 1.1262)

Example 2: Approximate (0.99)⁵

Write as (1 − 0.01)⁵

≈ 1 + 5(−0.01) + 10(−0.01)²

≈ 1 − 0.05 + 0.001 = 0.951

Rule of thumb: Keep terms up to x² or x³ for sufficient accuracy when |x| is small.

📊 Patterns in Binomial Expansion

Key Observations

Property	Description
Number of terms	(a+b)ⁿ has (n+1) terms
Powers of a	Decrease from n to 0
Powers of b	Increase from 0 to n
Sum of powers	Always equals n
Coefficients	Symmetric (palindrome)
Sum of coefficients	2ⁿ (set a=b=1)

Verification

Sum of coefficients of (1+x)⁴:

Coefficients: 1+4+6+4+1 = 16 = 2⁴ ✓

Number of terms: 5 = 4+1 ✓

💡 Real-Life Applications

Probability (Binomial Distribution)

Example:

A coin is tossed 4 times. What is the probability of getting exactly 2 heads?

P(2 heads) = ⁴C₂ × (0.5)² × (0.5)² = 6 × 0.25 × 0.25 = 0.375

Compound Interest Approximation

Example:

Approximate (1.05)⁴ using binomial (rate = 5% for 4 years)

(1+0.05)⁴ ≈ 1 + 4(0.05) + 6(0.05)² + 4(0.05)³

≈ 1 + 0.2 + 0.015 + 0.0005 = 1.2155

Actual: 1.2155 ✓

Population Growth

Example:

Population grows at 3% per year. Approximate population after 5 years if current is 1 million.

(1.03)⁵ ≈ 1 + 5(0.03) + 10(0.03)² = 1 + 0.15 + 0.009 = 1.159

Population ≈ 1,159,000

✏️ Practice Problems

Problem 1

Expand fully (2x − 1)⁴.

Solution:

a=2x, b=−1, n=4. Coefficients: 1,4,6,4,1

= (2x)⁴ + 4(2x)³(−1) + 6(2x)²(−1)² + 4(2x)(−1)³ + (−1)⁴

= 16x⁴ − 32x³ + 24x² − 8x + 1

Problem 2

Find the coefficient of x³ in the expansion of (3 + 2x)⁵.

Solution:

T₄ = ⁵C₃(3)²(2x)³ = 10 × 9 × 8x³ = 720x³

Coefficient = 720

Problem 3

Use binomial expansion to evaluate (1.01)⁵ to 4 decimal places.

Solution:

(1+0.01)⁵ ≈ 1+5(0.01)+10(0.01)²+10(0.01)³

= 1+0.05+0.001+0.00001 = 1.0510

📝 Quiz — Binomial Expansion

Part A: Fill in the Blanks (5 marks)

1. The expansion of (a+b)ⁿ has terms.

2. ⁶C₂ = .

3. The general term formula is Tᵣ₊₁ = ⁿCᵣ · .

4. The sum of all coefficients in (1+x)⁵ is .

5. The 3rd row of Pascal's Triangle (for n=2) gives coefficients .

Part B: Short Answer (10 marks)

6. (4 marks) Expand (1 + 2x)⁴ fully and find the term independent of x.

7. (3 marks) Find the coefficient of x³ in the expansion of (2−x)⁶.

8. (3 marks) Use binomial expansion to evaluate (0.98)⁴ to 4 decimal places.