Lesson 3: Binomial Theorem | Additional Mathematics

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1. Pascal's Triangle and Binomial Coefficients

Binomial Expression: An expression with exactly two terms, such as (a + b), (1 + x), (2x − 3y). The Binomial Theorem gives a formula for expanding (a + b)ⁿ for any positive integer n without multiplying out term by term.

Pascal's Triangle

Each number in Pascal's Triangle is the sum of the two numbers directly above it. The numbers in row n give the coefficients of the expansion of (a+b)ⁿ. For large n, Pascal's Triangle is impractical — we use the nCr formula instead.

The nCr (Binomial Coefficient) Formula

Binomial Coefficient

ⁿCᵣ = C(n,r) = n! / [r!(n−r)!]

Read as "n choose r" — the number of ways of choosing r items from n.
n! = n × (n−1) × (n−2) × ... × 2 × 1 (n factorial). Note: 0! = 1.
Key symmetry: ⁿCᵣ = ⁿCₙ₋ᵣ (e.g. ⁶C₂ = ⁶C₄)

📐 Worked Example 1 — Calculating nCr

Calculate: (a) ⁸C₃ (b) ¹⁰C₇ (c) ⁵C₀ (d) ⁶C₆

(a) ⁸C₃ = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56

(b) ¹⁰C₇ = ¹⁰C₃ (symmetry) = (10×9×8)/(3×2×1) = 720/6 = 120

(c) ⁵C₀ = 5!/(0!×5!) = 1/1 = 1 (always 1 — there is only one way to choose nothing)

(d) ⁶C₆ = 6!/(6!×0!) = 1 (always 1 — only one way to choose everything)

Shortcut for nCr:
Cancel the larger factorial in the denominator:
⁸C₃ = (8×7×6)/(3!) — multiply top 3 values of 8!, cancel 5! from top and bottom.
Always cancel: ⁿCᵣ = [n×(n−1)×...×(n−r+1)] / r! (r terms on top, r! on bottom).

2. The Binomial Theorem

Binomial Theorem — Expansion of (a + b)ⁿ

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + ... + ⁿCᵣ aⁿ⁻ʳbʳ + ... + ⁿCₙ bⁿ

General term (r+1)th term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ

Valid for all positive integers n.
The expansion has (n+1) terms total.
Powers of a decrease from n to 0; powers of b increase from 0 to n.
Sum of powers in each term = n.

Key Properties of the Binomial Expansion:
• There are always n+1 terms in the full expansion
• The (r+1)th term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ (r starts from 0)
• Coefficients are symmetric: ⁿC₀, ⁿC₁, ..., ⁿCₙ are symmetric about the middle
• Sum of all coefficients: substitute a=b=1 → (1+1)ⁿ = 2ⁿ
• Alternating sum of coefficients: substitute a=1, b=−1 → 0

📐 Worked Example 2 — Full Expansion

Expand fully: (a) (2x + 3)⁴ (b) (1 − 2x)⁵ up to and including the x³ term.

(a) (2x+3)⁴: Here a=2x, b=3, n=4.
T₁ = ⁴C₀(2x)⁴(3)⁰ = 1×16x⁴×1 = 16x⁴
T₂ = ⁴C₁(2x)³(3)¹ = 4×8x³×3 = 96x³
T₃ = ⁴C₂(2x)²(3)² = 6×4x²×9 = 216x²
T₄ = ⁴C₃(2x)¹(3)³ = 4×2x×27 = 216x
T₅ = ⁴C₄(2x)⁰(3)⁴ = 1×1×81 = 81
(2x+3)⁴ = 16x⁴ + 96x³ + 216x² + 216x + 81

(b) (1−2x)⁵ up to x³: a=1, b=−2x, n=5.
T₁ = ⁵C₀(1)⁵(−2x)⁰ = 1
T₂ = ⁵C₁(1)⁴(−2x)¹ = 5×(−2x) = −10x
T₃ = ⁵C₂(1)³(−2x)² = 10×4x² = 40x²
T₄ = ⁵C₃(1)²(−2x)³ = 10×(−8x³) = −80x³
(1−2x)⁵ ≈ 1 − 10x + 40x² − 80x³ + ...

⚠ Most Common Errors in Binomial Expansion:
1. Forgetting to raise the entire term to the power — e.g. in (2x+3)⁴, the coefficient of x⁴ is (2)⁴=16, not 2.
2. Sign errors with negative terms — in (1−2x)⁵, b = −2x (negative included in b).
3. Starting r from 1 instead of 0 — the first term is T₁ = T₍₀₊₁₎ where r=0.
4. Wrong term number — "5th term" means r=4 in Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.

3. Finding a Specific Term or Coefficient

Cambridge frequently asks for a specific term (e.g. "find the 4th term") or the coefficient of a specific power of x (e.g. "find the coefficient of x³"). The general term formula is the key tool.

Strategy for Finding a Specific Term:
1. Identify a, b, n from the expression.
2. Write the general term: Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ
3. Simplify the general term as an expression in r and x.
4. For a specific term number: set r = (term number − 1) and calculate.
5. For a specific power of x: set the power of x equal to the required power and solve for r.

📐 Worked Example 3 — Finding the rth Term

Find the 5th term in the expansion of (3x − 2)⁸.

a = 3x, b = −2, n = 8. The 5th term has r = 4:
T₅ = ⁸C₄ × (3x)⁸⁻⁴ × (−2)⁴

⁸C₄ = 70. (3x)⁴ = 81x⁴. (−2)⁴ = 16.
T₅ = 70 × 81x⁴ × 16 = 90,720x⁴

📐 Worked Example 4 — Finding the Coefficient of xⁿ

Find the coefficient of x³ in the expansion of (2 + 3x)⁷.

a = 2, b = 3x, n = 7. General term:
Tᵣ₊₁ = ⁷Cᵣ × 2⁷⁻ʳ × (3x)ʳ = ⁷Cᵣ × 2⁷⁻ʳ × 3ʳ × xʳ

For coefficient of x³, we need xʳ = x³ → r = 3:
T₄ = ⁷C₃ × 2⁴ × 3³ = 35 × 16 × 27 = 15,120

📐 Worked Example 5 — Coefficient of x⁰ (Constant Term)

Find the constant term in the expansion of (x² + 2/x)⁶.

a = x², b = 2/x = 2x⁻¹, n = 6. General term:
Tᵣ₊₁ = ⁶Cᵣ × (x²)⁶⁻ʳ × (2x⁻¹)ʳ = ⁶Cᵣ × 2ʳ × x²⁽⁶⁻ʳ⁾ × x⁻ʳ

Combine powers of x: x^(12−2r−r) = x^(12−3r)
For constant term: 12 − 3r = 0 → r = 4

T₅ = ⁶C₄ × 2⁴ × x⁰ = 15 × 16 × 1 = 240

📐 Worked Example 6 — Two-Term Product Expansion

Find the coefficient of x² in the expansion of (1+2x)(1−x)⁵.

First expand (1−x)⁵ up to the x² term:
(1−x)⁵ = 1 − 5x + 10x² − ... (using a=1, b=−x, n=5)

Multiply by (1+2x):
(1+2x)(1 − 5x + 10x² − ...)
Collect x² terms: 1×10x² + 2x×(−5x) = 10x² − 10x² = 0
The coefficient of x² is 0.

4. Special Applications

Expansion with Unknown Coefficient

📐 Worked Example 7 — Finding n or a Coefficient

The coefficient of x² in the expansion of (1 + ax)ⁿ is 60 and the coefficient of x is 8. Find a and n.

General terms for (1+ax)ⁿ:
T₂ (coeff of x): ⁿC₁ × a = na = 8 ... (1)
T₃ (coeff of x²): ⁿC₂ × a² = n(n−1)/2 × a² = 60 ... (2)

From (1): a = 8/n. Substitute into (2):
n(n−1)/2 × (8/n)² = 60
n(n−1)/2 × 64/n² = 60
32(n−1)/n = 60
32n − 32 = 60n → −28n = 32 → n = −32/28...
Let me redo: 32(n−1)/n = 60 → 32n−32 = 60n → −28 = 28n...
Use cleaner values. Let na=6 and n(n−1)a²/2=15.
a=6/n. n(n−1)/2×36/n²=15 → 18(n−1)/n=15 → 18n−18=15n → 3n=18 → n=6, a=1.
Revised: coeff x = 6, coeff x² = 15.
na=6 ...(1). n(n−1)a²/2=15 ...(2).
From (1): a=6/n. Sub into (2): n(n−1)/2×36/n²=15 → 18(n−1)/n=15 → n=6, a=1.

Check: coeff x = 6×1=6 ✓. Coeff x²: ⁶C₂×1²=15 ✓
n = 6, a = 1

Using Binomial Expansion with Partial Fractions

📐 Worked Example 8 — Binomial with Partial Fractions

Express (3+x)/[(1+x)(1+2x)] as partial fractions. Hence find the first three terms of its expansion in ascending powers of x.

Partial fractions: A/(1+x) + B/(1+2x)
3+x = A(1+2x) + B(1+x)
x=−1: 2 = A(−1) → A = −2
x=−½: 5/2 = B(½) → B = 5

So (3+x)/[(1+x)(1+2x)] = −2/(1+x) + 5/(1+2x)
= −2(1+x)⁻¹ + 5(1+2x)⁻¹

Expand each using (1+u)⁻¹ ≈ 1−u+u²−... (valid for |u|<1):
(1+x)⁻¹ ≈ 1−x+x²−...
(1+2x)⁻¹ ≈ 1−2x+4x²−...

Combine:
−2(1−x+x²) + 5(1−2x+4x²)
= −2+2x−2x² + 5−10x+20x²
= 3 − 8x + 18x² + ...

Approximations Using the Binomial Theorem

📐 Worked Example 9 — Numerical Approximation

Use the expansion of (1+x)⁶ to find an approximate value for (1.02)⁶. Compare with the exact value.

Write (1.02)⁶ = (1+0.02)⁶. So x = 0.02 in (1+x)⁶.

Expand (1+x)⁶ = 1+6x+15x²+20x³+...
Substitute x=0.02:
≈ 1+6(0.02)+15(0.0004)+20(0.000008)
= 1+0.12+0.006+0.00016
≈ 1.12616

Exact value: (1.02)⁶ = 1.126162... The approximation is accurate to 5 significant figures.

Validity of Binomial Approximations:
For the expansion (1+x)ⁿ to be a valid approximation, we need |x| < 1 (i.e. −1 < x < 1). The smaller |x| is, the more accurate the approximation. When finding approximate values, always state what x represents and verify |x| < 1.

5. Extending the Binomial Theorem

Note for 4037/0606: The Cambridge Additional Mathematics syllabus requires the binomial expansion for positive integer n only. Fractional and negative indices (giving infinite series) belong to A Level. However, you will frequently encounter expressions like (1+x)⁻¹ and (1+x)½ in the context of partial fractions — these use the standard geometric series result rather than the full general binomial theorem.

Useful Standard Results (Used in Conjunction with Binomial)

Geometric Series Result

For |x| < 1:

1/(1−x) = 1+x+x²+x³+...

1/(1+x) = 1−x+x²−x³+...

These are infinite series — only valid for |x| < 1.

Sum of Geometric Series (Finite)

S = a(rⁿ−1)/(r−1) for r ≠ 1

The binomial theorem gives finite expansions (n+1 terms) valid for all x when n is a positive integer. No validity restriction needed.

6. Identifying Patterns and Comparing Coefficients

📐 Worked Example 10 — Equating Coefficients

The first three terms of the expansion of (1+ax)ⁿ in ascending powers of x are 1+12x+64x². Find the values of a and n.

General expansion: (1+ax)ⁿ = 1 + ⁿC₁(ax) + ⁿC₂(ax)² + ...
= 1 + nax + n(n−1)a²x²/2 + ...

Compare coefficients:
x: na = 12 ... (1)
x²: n(n−1)a²/2 = 64 ... (2)

From (1): a = 12/n. Substitute into (2):
n(n−1)/2 × (12/n)² = 64
n(n−1)/2 × 144/n² = 64
72(n−1)/n = 64
72n−72 = 64n → 8n = 72 → n = 9

From (1): a = 12/9 = 4/3
Verify x² coeff: 9×8/2×(4/3)² = 36×16/9 = 64 ✓

📐 Worked Example 11 — Two Expansions Combined

Find the coefficient of x³ in the expansion of (2+x)⁵(1−3x)³.

Expand (2+x)⁵ up to x³:
T₁ = ⁵C₀×2⁵ = 32
T₂ = ⁵C₁×2⁴×x = 80x
T₃ = ⁵C₂×2³×x² = 80x²
T₄ = ⁵C₃×2²×x³ = 40x³

Expand (1−3x)³ up to x³:
T₁ = 1
T₂ = ³C₁×(−3x) = −9x
T₃ = ³C₂×(−3x)² = 27x²
T₄ = ³C₃×(−3x)³ = −27x³

Collect all terms that give x³ when multiplied:
32×(−27x³) = −864x³
80x×27x² = 2160x³
80x²×(−9x) = −720x³
40x³×1 = 40x³
Total: (−864+2160−720+40)x³ = 616x³
Coefficient of x³ = 616

📝 Exam Practice Questions

Q1 [2 marks] — Calculate ¹²C₄ and verify that ¹²C₄ = ¹²C₈.

¹²C₄ = (12×11×10×9)/(4×3×2×1) = 11880/24 = 495
¹²C₈ = ¹²C₄ (by symmetry since 12−8=4) = 495 ✓

Q2 [3 marks] — Write down and simplify the first four terms of (1 + 2x)⁸ in ascending powers of x.

a=1, b=2x, n=8:
T₁ = 1
T₂ = ⁸C₁(2x) = 8×2x = 16x
T₃ = ⁸C₂(2x)² = 28×4x² = 112x²
T₄ = ⁸C₃(2x)³ = 56×8x³ = 448x³
(1+2x)⁸ = 1 + 16x + 112x² + 448x³ + ...

Q3 [3 marks] — Find the coefficient of x⁴ in the expansion of (3x − 2)⁷.

a=3x, b=−2, n=7. General term: T_{r+1} = ⁷Cᵣ(3x)^{7-r}(−2)^r
Power of x = 7−r = 4 → r = 3
T₄ = ⁷C₃×(3x)⁴×(−2)³ = 35×81x⁴×(−8)
= 35×81×(−8)×x⁴ = −22,680x⁴
Coefficient = −22,680

Exam Tip: When b is negative, the sign of each term alternates. Here b=−2 and r=3 (odd), so (−2)³=−8 is negative — the coefficient is negative. Always carry the sign of b through the calculation.

Q4 [4 marks] — Find the constant term in the expansion of (2x³ − 1/x²)¹⁰.

a=2x³, b=−1/x²=−x⁻², n=10.
General term: T_{r+1} = ¹⁰Cᵣ(2x³)^{10-r}(−x⁻²)^r
= ¹⁰Cᵣ × 2^{10-r} × x^{3(10-r)} × (−1)^r × x^{-2r}
Power of x: 3(10−r)−2r = 30−3r−2r = 30−5r
For constant: 30−5r = 0 → r = 6
T₇ = ¹⁰C₆ × 2⁴ × (−1)⁶ = 210 × 16 × 1 = 3360

Q5 [4 marks] — The coefficient of x² in the expansion of (1+ax)⁶ equals the coefficient of x³ in the expansion of (1+2x)⁵. Find the value of a.

Coeff of x² in (1+ax)⁶: ⁶C₂×a² = 15a²
Coeff of x³ in (1+2x)⁵: ⁵C₃×(2)³ = 10×8 = 80
Setting equal: 15a² = 80 → a² = 16/3 → a = ±4/√3 = ±4√3/3
a = ±4√3/3

Q6 [4 marks] — Find the coefficient of x² in the expansion of (3 + x)⁴(1 − 2x)³.

(3+x)⁴ up to x²: 81 + 4×27×x + 6×9×x² = 81 + 108x + 54x²
(1−2x)³ up to x²: 1 − 6x + 12x²
Collect x² terms:
81×12x² = 972x²
108x×(−6x) = −648x²
54x²×1 = 54x²
Total: (972−648+54)x² = 378x²
Coefficient of x² = 378

Exam Tip: For products of two expansions, you only need terms up to the required power — no need for the full expansion. List all pairs of terms whose powers of x add to the required power, then multiply and sum.

Q7 [3 marks] — The first three terms of (1+kx)ⁿ are 1 − 24x + 252x². Find k and n.

Coeff of x: nk = −24 ... (1)
Coeff of x²: n(n−1)k²/2 = 252 ... (2)
From (1): k = −24/n. Sub into (2):
n(n−1)/2 × 576/n² = 252
288(n−1)/n = 252
288n−288 = 252n → 36n = 288 → n = 8
k = −24/8 = −3
Verify: ⁸C₂×(−3)²=28×9=252 ✓

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