Binomial Theorem: Expansion, General Term and Middle Term

Expanding powers of a sum

You already know (a + b)² = a² + 2ab + b² and (a + b)³ = a³ + 3a²b + 3ab² + b³. Multiplying out (a + b)¹⁰ by hand would take a long time. The binomial theorem gives the answer directly.

Pascal's triangle

Write the coefficients of successive powers in rows:

n = 0              1
n = 1            1   1
n = 2          1   2   1
n = 3        1   3   3   1
n = 4      1   4   6   4   1
n = 5    1   5  10  10   5   1

Each row starts and ends with 1, and every other number is the sum of the two numbers above it. The row for n has n + 1 entries, which are the combinations ⁿC₀, ⁿC₁, …, ⁿCₙ.

Recall ⁿCᵣ = n! ⁄ (r!(n − r)!), with ⁿC₀ = ⁿCₙ = 1 and ⁿCᵣ = ⁿCₙ₋ᵣ.

The binomial theorem

For any positive integer n:

(a + b)ⁿ = ⁿC₀ aⁿ + ⁿC₁ aⁿ⁻¹b + ⁿC₂ aⁿ⁻²b² + … + ⁿCₙ bⁿ

Observations:

There are n + 1 terms.
The power of a decreases by 1 from n to 0; the power of b increases from 0 to n.
In every term, the powers of a and b add up to n.
Coefficients equidistant from the ends are equal.

Example. (x + 2)⁴ = x⁴ + 4x³(2) + 6x²(4) + 4x(8) + 16 = x⁴ + 8x³ + 24x² + 32x + 16.

Special cases:

(a − b)ⁿ: the signs alternate, + − + − …
(1 + x)ⁿ = 1 + ⁿC₁x + ⁿC₂x² + … + xⁿ
Putting a = b = 1: ⁿC₀ + ⁿC₁ + … + ⁿCₙ = 2ⁿ. The coefficients in any row of Pascal's triangle add up to a power of 2.

The general term

The (r + 1)th term of (a + b)ⁿ is

Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ

Notice the index: the 5th term uses r = 4.

Example. Find the 4th term of (2x − 3)⁶. Here r = 3:

T₄ = ⁶C₃ (2x)³ (−3)³ = 20 × 8x³ × (−27) = −4320x³.

Finding a particular coefficient

Write the general term, collect the power of x, and set it equal to the required power.

Example. Find the coefficient of x⁵ in (x + 3)⁸.

Tᵣ₊₁ = ⁸Cᵣ x⁸⁻ʳ 3ʳ. For x⁵, 8 − r = 5, so r = 3. The coefficient is ⁸C₃ × 3³ = 56 × 27 = 1512.

Term independent of x. Find the term free of x in (x² + 1/x)⁶.

Tᵣ₊₁ = ⁶Cᵣ (x²)⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x¹²⁻³ʳ. Setting 12 − 3r = 0 gives r = 4, so the term is ⁶C₄ = 15.

Middle terms

If n is even, there is one middle term: the (n/2 + 1)th.
If n is odd, there are two middle terms: the ((n + 1)/2)th and ((n + 3)/2)th.

Example. The middle term of (x + 1/x)¹⁰ is the 6th (r = 5): ¹⁰C₅ x⁵ (1/x)⁵ = 252.

Example. (a + b)⁷ has two middle terms, the 4th and 5th: 35a⁴b³ and 35a³b⁴.

The middle term has the largest binomial coefficient.

Quick calculations

Example. Find (1.02)⁵ to four decimal places.

(1 + 0.02)⁵ = 1 + 5(0.02) + 10(0.0004) + 10(0.000008) + … = 1 + 0.1 + 0.004 + 0.00008 + … ≈ 1.1041.

Example. Which is larger, (1.1)¹⁰⁰⁰⁰ or 1000?

(1 + 0.1)¹⁰⁰⁰⁰ = 1 + 10000(0.1) + (positive terms) = 1 + 1000 + … > 1000. So (1.1)¹⁰⁰⁰⁰ is larger.

Divisibility. Show that 9ⁿ − 8n − 1 is divisible by 64. Write 9ⁿ = (1 + 8)ⁿ = 1 + 8n + ⁿC₂ 8² + ⁿC₃ 8³ + …. Then 9ⁿ − 8n − 1 = 64(ⁿC₂ + ⁿC₃ 8 + …), a multiple of 64.

Summary

(a + b)ⁿ = Σ ⁿCᵣ aⁿ⁻ʳ bʳ, with n + 1 terms whose powers add up to n.
General term: Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ.
Sum of the coefficients is 2ⁿ.
Middle term: (n/2 + 1)th for even n; two middle terms for odd n.