Pascal's triangle and the binomial theorem

You already know (x + y)² and (x + y)³. This chapter lets you open any power of a two-term expression (a binomial), even (x + y)¹¹, without multiplying it out bracket by bracket.

Look at the first few powers and three patterns appear. The power n is called the index.

• Number of terms = index + 1. (x + y)² has 3 terms, (x + y)³ has 4, so (x + y)¹⁰ will have 11.
• Powers slide. The power of the first term starts at n and falls by 1 each time down to 0; the power of the second term starts at 0 and rises to n.
• Every term has total degree n. In (x + y)³ = x³ + 3x²y + 3xy² + y³, each term's powers add up to 3 (x²y is 2 + 1).

The numbers in front (the coefficients) come from Pascal's triangle. Each row starts and ends with 1, and every inside number is the sum of the two numbers just above it. In India the same triangle was known long before, as Pingala's Meru Prastara.

(The row for 6 comes from the row for 5 by adding neighbours: you can build it yourself.)

Example: (x + 2y)⁴ uses the row 1, 4, 6, 4, 1. Write the powers of x falling from 4 and the powers of 2y rising from 0:

• 1·x⁴ + 4·x³(2y) + 6·x²(2y)² + 4·x(2y)³ + 1·(2y)⁴
• = x⁴ + 8x³y + 24x²y² + 32xy³ + 16y⁴.

Building rows gets slow for big powers, so we use combinations. The coefficients of (x + y)ⁿ are exactly ⁿC₀, ⁿC₁, ⁿC₂, ..., ⁿCₙ. That is the binomial theorem:

• (x + y)ⁿ = ⁿC₀xⁿ + ⁿC₁xⁿ⁻¹y + ⁿC₂xⁿ⁻²y² + ... + ⁿCₙyⁿ.
• Facts that make it quick: ⁿC₀ = ⁿCₙ = 1, ⁿC₁ = n, and ⁿCᵣ = ⁿCₙ₋ᵣ (so ⁸C₆ = ⁸C₂ = 28). The coefficients read the same from both ends.

A minus sign in the middle. Write x – y as x + (–y). Odd powers of –y stay negative and even powers turn positive, so the signs simply alternate + – + – ..., and the last term carries (–1)ⁿ.

• (a – 3b)³ = a³ – 3a²(3b) + 3a(3b)² – (3b)³ = a³ – 9a²b + 27ab² – 27b³.

Putting x = y = 1 turns the theorem into a fact about the coefficients: ⁿC₀ + ⁿC₁ + ... + ⁿCₙ = 2ⁿ. For n = 5: 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵. With x = 1, y = –1 the alternating sum is 0.