The Binomial Theorem
“The binomial theorem's coefficients are just combinations — Pascal's triangle.”
The formula
(a + b)ⁿ = Σₖ₌₀ⁿ ₙCₖ aⁿ⁻ᵏ bᵏHow to read it: expanding (a+b) multiplied by itself n times makes each term's coefficient a combination ₙCₖ
- ₙCₖ
- — the binomial coefficient — the number in front of the aⁿ⁻ᵏbᵏ term
- aⁿ⁻ᵏ bᵏ
- — a term with a used n−k times and b used k times
- n
- — how many times the bracket is multiplied (the power)
The hook
You don't have to multiply out (a+b)¹⁰ by hand — once you know each term's coefficient is a 'combination', you can expand it in one line.
In plain words
When you expand (a+b)ⁿ, each term's coefficient is ₙCₖ — the number of ways to choose b from k of the n brackets.
The intuition
(a+b)ⁿ is (a+b) written out n times and multiplied. Expanding means picking one of a or b from each bracket and adding up all the products. The number of ways b gets chosen exactly k times is ₙCₖ → so the coefficient of aⁿ⁻ᵏbᵏ is ₙCₖ. The coefficients were 'counts of ways to choose' all along.
How it's built
For a fixed power n, the terms are aⁿ, aⁿ⁻¹b, …, bⁿ, the exponents always summing to n. The coefficients ₙC₀, ₙC₁, …, ₙCₙ are exactly one row of Pascal's triangle — the one where each entry is the sum of the two above.
Example
(a+b)² = ₂C₀a² + ₂C₁ab + ₂C₂b² = 1·a² + 2ab + 1·b². The coefficients 1, 2, 1 are the third row of Pascal's triangle.
Common mistake
(a+b)ⁿ ≠ aⁿ + bⁿ. The middle cross-terms, carrying combination coefficients, always appear. Even (a+b)² is wrong if you drop the 2ab.
Where it's used
Computing the binomial distribution in probability, approximating (1+x)ⁿ, expanding algebraic expressions, combinatorial identities — a bridge between probability/statistics and algebra.
Where it came from
Arranging the coefficients as 'Pascal's triangle' was known before Pascal in China (Yang Hui) and Persia — wisdom that crossed civilizations.
Prerequisites
Quick check
In the expansion of (a+b)³, what is the coefficient of a²b?
- 1
- 2
- 3✓
- 6
Practice
Find the coefficient of a²b² in the expansion of (a+b)⁴.
Answer: 6
- Ways b is chosen twice: ₄C₂
- = (4·3)/(2·1) = 6
Takeaway: The coefficient is ₙCₖ, where k is b's exponent.
Find the coefficient of x² in the expansion of (1+x)⁵.
Answer: 10
- Ways x is chosen twice: ₅C₂
- = (5·4)/(2·1) = 10
Takeaway: In (1+x)ⁿ the coefficient of xᵏ is ₙCₖ.
Find the sum of all coefficients in (a+b)⁴. (Hint: set a=b=1)
Answer: 16
- Substitute a=b=1: (1+1)⁴
- = 2⁴ = 16
Takeaway: The sum of coefficients is a=b=1 → 2ⁿ.
Explain how the coefficients 1, 3, 3, 1 of (a+b)³ connect to Pascal's triangle.
Answer: undefined
- In Pascal's triangle each entry is the sum of the two above
- Adding neighbours of 1,2,1 gives 1,3,3,1
- That row is exactly the coefficients of (a+b)³
Takeaway: Binomial coefficients are one row of Pascal's triangle.