04 // Interactive Formula Compendium // CALCULUS-ALGEBRA(a + b)ⁿ = Σ_{k=0..n} C(n, k) · aⁿ⁻ᵏ · bᵏ
Binomial Theorem Formula — Expansion, Pascal's Triangle & General Term
Newton's Binomial Theorem states that for any non-negative integer n, the polynomial expansion of (a + b)ⁿ is the sum from k = 0 to n of C(n, k) aⁿ⁻ᵏ bᵏ, where C(n, k) = n! / (k!(n − k)!).
Formula Reference
Main Formula
(a + b)ⁿ = Σ_{k=0..n} C(n, k) · aⁿ⁻ᵏ · bᵏ
Rearranged Forms
T_{k+1} = C(n, k) · aⁿ⁻ᵏ · bᵏ (General k-th Indexed Term)
C(n, k) = n! / (k! (n − k)!) (Binomial Coefficient)
Σ_{k=0..n} C(n, k) = 2ⁿ (Sum of Pascal Row)
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| T_{k+1} | Evaluated Binomial Term C(n,k) aⁿ⁻ᵏ bᵏ | dimensionless |
| a | First Base Term (a) | 1 |
| b | Second Base Term (b) | 1 |
| n | Binomial Exponent (n) | integer |
| k | Term Index (0 ≤ k ≤ n) | integer |
Inputs & Parameters
[1]
[1]
[integer]
[integer]
Calculated Result (Evaluated Binomial Term C(n,k) aⁿ⁻ᵏ bᵏ)
T_{k+1} =216dimensionless
Formula Notes & Background
- Multiplying n identical factors (a + b)(a + b)...(a + b) produces terms of the form aⁿ⁻ᵏbᵏ by choosing 'b' from exactly k factors and 'a' from the remaining n − k factors.
- The number of distinct ways to select k factors out of n is the combinatorial coefficient C(n, k) = n! / (k!(n − k)!).
- Pascal's identity C(n, k) + C(n, k−1) = C(n+1, k) enables inductive proof for all integers n ≥ 0.
Step-by-Step Examples
#1: Find the k = 2 middle term in the expansion of (2 + 3)⁴.
a = 2b = 3n = 4k = 2
T₃ = C(4, 2) × 2^(4−2) × 3² = 6 × 4 × 9
→ T₃ = 216 (while full expansion (2+3)⁴ = 5⁴ = 625)
#2: Find the coefficient of x³y² in (2x + y)⁵.
a = 2b = 1n = 5k = 2
Coeff = C(5, 2) × 2³ × 1² = 10 × 8 × 1
→ 80 (so the term is 80x³y²)
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Frequently Asked Questions
How many total terms are in the expansion of (a + b)ⁿ?+
There are exactly n + 1 terms, indexed from k = 0 (which gives aⁿ) to k = n (which gives bⁿ).
Can the Binomial Theorem be used for fractional or negative exponents?+
Yes. Generalized Newton binomial series expands (1 + x)^α = Σ [α(α−1)...(α−k+1)/k!] xᵏ, which converges for |x| < 1.
Why do the binomial coefficients in row n sum to 2ⁿ?+
Setting a = 1 and b = 1 in (a + b)ⁿ = Σ C(n, k) aⁿ⁻ᵏ bᵏ immediately yields (1 + 1)ⁿ = 2ⁿ = Σ C(n, k).