Pascal's Triangle & Binomial Coefficients
📌 What is a Binomial?
An expression with two terms connected by + or − is a binomial, e.g. (a + b), (x − y). The Binomial Theorem gives a quick method to expand (a + b)ⁿ for any positive integer n without repeated multiplication.
📌 Key Observations in Binomial Expansion
(i) Total number of terms in (a + b)ⁿ = n + 1
(ii) Powers of 'a' decrease by 1 each term; powers of 'b' increase by 1.
(iii) In every term, sum of indices of a and b = n.
(iv) Coefficients follow Pascal's Triangle pattern.
🔺 Pascal's Triangle (Coefficients)
| Index 0 | 1 | |||||
| Index 1 | 1 | 1 | ||||
| Index 2 | 1 | 2 | 1 | |||
| Index 3 | 1 | 3 | 3 | 1 | ||
| Index 4 | 1 | 4 | 6 | 4 | 1 | |
| Index 5 | 1 | 5 | 10 | 10 | 5 | 1 |
Example (index 4): ⁴C₀=1, ⁴C₁=4, ⁴C₂=6, ⁴C₃=4, ⁴C₄=1
Each element = sum of two elements directly above it.
🎯 Practice MCQs
Number of terms in (a + b)ⁿ = n + 1 = 10 + 1 = 11.
Row for index 5: 1, 5, 10, 10, 5, 1. Coefficient of x³y² (r=2) = ⁵C₂ = 10.
Sum of binomial coefficients = 2ⁿ = 2⁷ = 128.
Binomial Theorem — General Statement
📌 The Binomial Theorem (Positive Integer n)
For any positive integer n and any real numbers a and b:
Compact form: (a + b)ⁿ = Σ ⁿCₖ aⁿ⁻ᵏ bᵏ (k = 0 to n)
Proved by Principle of Mathematical Induction
📌 Worked Example: (x + 2)⁶
(x + 2)⁶ = ⁶C₀x⁶ + ⁶C₁x⁵·2 + ⁶C₂x⁴·4 + ⁶C₃x³·8 + ⁶C₄x²·16 + ⁶C₅x·32 + ⁶C₆·64
= x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64
🎯 Practice MCQs
(100−2)³ = 100³ − 3·100²·2 + 3·100·4 − 8
= 1000000 − 60000 + 1200 − 8 = 941192.
Term with x⁴: r = 3, so ⁷C₃ · x⁴ · 3³ = 35 × 27 = 945.
Wait — ⁷C₃ · (3)³ = 35 × 27 = 945. Answer is A) 945.
Putting x = 1 in (1+x)²⁰: Sum = 2²⁰ = 1048576.
Special Cases of Binomial Theorem
📌 Case 1: (x − y)ⁿ
Replace b with (−y):
(x − y)ⁿ = ⁿC₀xⁿ − ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² − ... + (−1)ⁿ ⁿCₙ yⁿ
Signs alternate: +, −, +, −, ... The r-th term has factor (−1)ʳ.
📌 Case 2: (1 + x)ⁿ
Put a = 1, b = x:
(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ... + ⁿCₙxⁿ
Special: x = 1 → 2ⁿ = ⁿC₀ + ⁿC₁ + ⁿC₂ + ... + ⁿCₙ
📌 Case 3: (1 − x)ⁿ
Put a = 1, b = −x:
(1 − x)ⁿ = ⁿC₀ − ⁿC₁x + ⁿC₂x² − ... + (−1)ⁿ ⁿCₙxⁿ
Special: x = 1 → 0 = ⁿC₀ − ⁿC₁ + ⁿC₂ − ... + (−1)ⁿ ⁿCₙ
Sum of all coefficients (all positive = 2ⁿ): ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿ
Result: 6ⁿ − 5n − 1 is always divisible by 25
🎯 Practice MCQs
(1+0.01)⁵ ≈ 1 + 5(0.01) + 10(0.01)² = 1 + 0.05 + 0.001 = 1.051 ≈ 1.0510.
This is (1−1)⁶ = 0. Alternating sum of binomial coefficients is always 0.
6ⁿ = (1+5)ⁿ = 1 + 5n + 25(ⁿC₂ + 5ⁿC₃ + ...)
∴ 6ⁿ − 5n = 1 + 25k. Remainder when divided by 25 = 1.
General Term & Applications
For (a + b)ⁿ: r = 0, 1, 2, ..., n
For (x − y)ⁿ: include (−1)ʳ → Tᵣ₊₁ = ⁿCᵣ · xⁿ⁻ʳ · (−y)ʳ
Middle term:
• If n is even → T(n/2)+1
• If n is odd → T(n+1)/2 and T(n+3)/2
📌 Finding a Specific Term
To find the term independent of x (constant term) in an expansion: write out general term Tᵣ₊₁, collect powers of x, set power = 0 and solve for r. Then substitute r back to get the value.
📌 Approximation Using Binomial Theorem
For small x, (1 + x)ⁿ ≈ 1 + nx (first two terms only). This is useful for approximating expressions like (0.99)⁵, (1.01)⁴ etc. without a calculator — a key skill for AAI ATC aptitude.
🎯 Practice MCQs
T₄ = T₃₊₁, r = 3: ⁸C₃ · x⁸⁻³ · 2³ = 56 × x⁵ × 8 = 448 x⁵.
Tᵣ₊₁ = ⁶Cᵣ · x⁶⁻ʳ · x⁻ʳ = ⁶Cᵣ · x⁶⁻²ʳ.
For constant term: 6 − 2r = 0 → r = 3.
T₄ = ⁶C₃ = 20.
(1−0.01)⁵ ≈ 1 − 5(0.01) + 10(0.01)²
= 1 − 0.05 + 0.001 = 0.951 ≈ 0.9510.
📚 Chapter Summary — Binomial Theorem
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