Binomial Theorem Expansion

[MathDefinition title="Binomial Theorem"] The binomial theorem provides a formula for expanding expressions of the form (a + b)ⁿ where n is a positive integer: (a + b)ⁿ = Σ(k=0 to n) ⁿCₖ aⁿ⁻ᵏ bᵏ [/MathDefinition]

[MathDefinition title="Binomial Expansion"] The expansion of (a + b)ⁿ has n + 1 terms: (a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + ... + ⁿCₙbⁿ [/MathDefinition]

[MathKeyFormula title="Key Formulas"]

1. (a + b)ⁿ = Σ(k=0 to n) ⁿCₖ aⁿ⁻ᵏ bᵏ
2. General term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ
3. Number of terms = n + 1
4. Sum of binomial coefficients: Σ ⁿCₖ = 2ⁿ
5. Alternating sum: Σ(−1)ᵏ ⁿCₖ = 0 [/MathKeyFormula]

[MathNote title="Properties of Binomial Expansion"]

1. The number of terms is n + 1
2. The sum of exponents in each term is n
3. The coefficients are symmetric: ⁿCₖ = ⁿCₙ₋ₖ
4. The middle term(s) have the largest coefficient(s) [/MathNote]

[MathExample title="Example 1: Basic Expansion"] Problem: Expand (x + y)⁴ using the binomial theorem.

Solution: Using (a + b)ⁿ = Σ(k=0 to n) ⁿCₖ aⁿ⁻ᵏ bᵏ

Here a = x, b = y, n = 4

(x + y)⁴ = ⁴C₀x⁴ + ⁴C₁x³y + ⁴C₂x²y² + ⁴C₃xy³ + ⁴C₄y⁴

⁴C₀ = 1, ⁴C₁ = 4, ⁴C₂ = 6, ⁴C₃ = 4, ⁴C₄ = 1

(x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴ [/MathExample]

[MathExample title="Example 2: Expansion with Negative Terms"] Problem: Expand (2x − 3y)³.

Solution: Using (a + b)ⁿ with a = 2x, b = −3y, n = 3

(2x − 3y)³ = ³C₀(2x)³ + ³C₁(2x)²(−3y) + ³C₂(2x)(−3y)² + ³C₃(−3y)³

³C₀ = 1, ³C₁ = 3, ³C₂ = 3, ³C₃ = 1

= 1(8x³) + 3(4x²)(−3y) + 3(2x)(9y²) + 1(−27y³) = 8x³ − 36x²y + 54xy² − 27y³ [/MathExample]

[MathExample title="Example 3: Finding Specific Term"] Problem: Find the 4th term in the expansion of (x + 2)⁶.

Solution: The general term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ

Here a = x, b = 2, n = 6

For the 4th term, r = 3: T₄ = T₃₊₁ = ⁶C₃ x⁶⁻³ (2)³

⁶C₃ = 6!/(3! × 3!) = 20

T₄ = 20 × x³ × 8 = 160x³

Therefore, the 4th term is 160x³. [/MathExample]

[MathExample title="Example 4: Finding Coefficient"] Problem: Find the coefficient of x³ in the expansion of (1 + x)⁸.

Solution: The general term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ

Here a = 1, b = x, n = 8

Tᵣ₊₁ = ⁸Cᵣ (1)⁸⁻ʳ xʳ = ⁸Cᵣ xʳ

For x³, we need r = 3: Coefficient = ⁸C₃ = 8!/(3! × 5!) = 56

Therefore, the coefficient of x³ is 56. [/MathExample]