Binomial Theorem for Positive Integers

History of the binomial theorem

The triangle of numbers behind this theorem is very old. In ancient India, Pingala (around the 2nd century BCE) counted patterns of long and short syllables in poetry, which leads to these numbers; later scholars such as Halayudha (10th century) described it as the Meru Prastara. Mathematicians in Persia (like Al-Karaji and Omar Khayyam) and China (Yang Hui) also knew it. In the 17th century Blaise Pascal studied its properties in detail, so we now call it Pascal's triangle. Isaac Newton later extended the idea to powers that are not whole numbers (not in this syllabus).

Statement of the binomial theorem

For any positive integer n and any numbers a and b:

(a + b)ⁿ = ⁿC₀aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + ⁿCₙ₋₁abⁿ⁻¹ + ⁿCₙbⁿ

• There are n + 1 terms.
• The power of a falls from n to 0; the power of b rises from 0 to n; in each term the powers add up to n.
• Coefficients are symmetric because ⁿCᵣ = ⁿCₙ₋ᵣ.
• Special cases: (a − b)ⁿ: put −b, so the signs alternate +, −, +, … (1 + x)ⁿ = ⁿC₀ + ⁿC₁x + … + ⁿCₙxⁿ.
• Putting x = 1 gives ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ.

Proof by mathematical induction

Let P(n) be the statement of the theorem.

1. Base case, n = 1: (a + b)¹ = ¹C₀a + ¹C₁b = a + b. True.
2. Assume P(k) is true: (a + b)ᵏ = ᵏC₀aᵏ + ᵏC₁aᵏ⁻¹b + … + ᵏCₖbᵏ.
3. Show P(k + 1): (a + b)ᵏ⁺¹ = (a + b)(a + b)ᵏ. Multiply each term by a and by b, then collect terms with the same powers. The term aᵏ⁺¹⁻ʳbʳ gets ᵏCᵣ (from a × ᵏCᵣaᵏ⁻ʳbʳ) plus ᵏCᵣ₋₁ (from b × ᵏCᵣ₋₁aᵏ⁻ʳ⁺¹bʳ⁻¹).
4. By Pascal's rule, ᵏCᵣ + ᵏCᵣ₋₁ = ᵏ⁺¹Cᵣ. The first and last coefficients are ᵏC₀ = ᵏ⁺¹C₀ = 1 and ᵏCₖ = ᵏ⁺¹Cₖ₊₁ = 1. So (a + b)ᵏ⁺¹ has exactly the form of P(k + 1).
5. By induction, P(n) is true for every positive integer n.

Step 5 of the 3D shows step 3 happening: each pillar of row 5 grows from its two neighbours in row 4.

Pascal's triangle

Write 1 at the top (row 0). Each row starts and ends with 1, and every inside number is the sum of the two numbers just above it.

Row 0: 1 · Row 1: 1 1 · Row 2: 1 2 1 · Row 3: 1 3 3 1 · Row 4: 1 4 6 4 1 · Row 5: 1 5 10 10 5 1 · Row 6: 1 6 15 20 15 6 1

Row n lists ⁿC₀, ⁿC₁, …, ⁿCₙ, the coefficients of (a + b)ⁿ. Each row is symmetric, and its sum is 2ⁿ. The triangle is quick for small n; for large n use the ⁿCᵣ formula.

Simple uses: general term, middle term, particular terms

• General term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ. For (a − b)ⁿ it is (−1)ʳ ⁿCᵣ aⁿ⁻ʳ bʳ.
• Middle term: if n is even, there is one middle term, T₍ₙ/₂₎₊₁. If n is odd, there are two: T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂.
• Term independent of x (constant term): write the power of x in Tᵣ₊₁, set it to 0, solve for r.
• Coefficient of xᵏ: set the power of x equal to k.
• Approximation and comparison: (0.99)⁵ = (1 − 0.01)⁵ ≈ 1 − 0.05 + 0.001 = 0.951. To compare 1.01¹⁰⁰⁰⁰⁰⁰ with 10000, expand and keep the first two terms: 1 + 1000000 × 0.01 = 10001 > 10000.

Board exam focus

Expect: expand a binomial like (2x − 3/x)⁴ (2–3 marks), find a general, middle or constant term (3 marks), evaluate (99)⁵ or (1.1)⁴ using the theorem (2 marks), and show that 9ⁿ⁺¹ − 8n − 9 is divisible by 64 (3 marks). Write Tᵣ₊₁ clearly before substituting.

Key formulas and definitions

• (a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, r = 0 to n (n + 1 terms)
• Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ; for (a − b)ⁿ: (−1)ʳ ⁿCᵣ aⁿ⁻ʳ bʳ
• Pascal's rule: ⁿCᵣ₋₁ + ⁿCᵣ = ⁿ⁺¹Cᵣ
• ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ
• Middle term: n even → T₍ₙ/₂₎₊₁; n odd → T₍ₙ₊₁₎/₂ and T₍ₙ₊₃₎/₂

Worked examples

Common mistakes

• Calling ⁿCᵣ aⁿ⁻ʳ bʳ the r-th term. It is the (r + 1)th term.
• Dropping the minus sign in (a − b)ⁿ. Put b = −y and keep the signs: they alternate.
• Forgetting to raise the coefficient too: (2x)³ = 8x³, not 2x³.
• Saying (a + b)ⁿ has n terms. It has n + 1 terms.