Binomial Expansion
Expand (a + b)ⁿ term by term, with exact coefficients for any power up to 30.
Coefficients sum to 32 = 2⁵
x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵
| k | C(n, k) | Powers | Term |
|---|---|---|---|
| 0 | 1 | 5 + 0 = 5 | x⁵ |
| 1 | 5 | 4 + 1 = 5 | 5x⁴y |
| 2 | 10 | 3 + 2 = 5 | 10x³y² |
| 3 | 10 | 2 + 3 = 5 | 10x²y³ |
| 4 | 5 | 1 + 4 = 5 | 5xy⁴ |
| 5 | 1 | 0 + 5 = 5 | y⁵ |
The binomial theorem expands (a + b)ⁿ without multiplying the brackets out. Each term is C(n, k)·aⁿ⁻ᵏ·bᵏ, and the powers always sum to n. So (x + y)⁵ = x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵.
Why the coefficients are what they are
Expanding (a + b)ⁿ means choosing, from each of the n brackets, either an a or a b. A term with exactly k copies of b arises once for every way of picking which k brackets contribute the b — and that count is C(n, k). The coefficients are not a pattern to memorise; they are the number of ways each term can happen.
That also explains Pascal’s triangle. Each entry is the sum of the two above it because choosing k items from n either includes the last item (leaving k − 1 from n − 1) or does not (leaving k from n − 1). The triangle is the recurrence C(n, k) = C(n−1, k−1) + C(n−1, k) drawn out.
Two patterns worth carrying
The expansion has n + 1 terms, and the powers of a and b always add to n — a quick check on any line you write. And setting a = b = 1 makes every term its own coefficient, so the coefficients of row n sum to 2ⁿ: 1 + 5 + 10 + 10 + 5 + 1 = 32 for n = 5.
C(n, k) = n! ÷ (k!(n − k)!); for (a − b)ⁿ the signs alternate, starting positive
- 1 Write out the coefficients. For n = 5 they are 1, 5, 10, 10, 5, 1 — row five of Pascal’s triangle.
- 2 Start the powers of a at n and count down. x⁵, x⁴, x³, x², x, then x⁰ = 1.
- 3 Start the powers of b at 0 and count up. 1, y, y², y³, y⁴, y⁵ — the two exponents always sum to 5.
- 4 Put them together. x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵.
- 5 Alternate the signs for a difference. (x − y)⁵ has the same coefficients with signs + − + − + −, since each odd power of −y is negative.
Pascal’s triangle to row 6
Each entry is the sum of the two above it. Row n gives the coefficients of (a + b)ⁿ.
| n | Coefficients | Sum = 2ⁿ |
|---|---|---|
| 0 | 1 | 1 |
| 1 | 1 1 | 2 |
| 2 | 1 2 1 | 4 |
| 3 | 1 3 3 1 | 8 |
| 4 | 1 4 6 4 1 | 16 |
| 5 | 1 5 10 10 5 1 | 32 |
| 6 | 1 6 15 20 15 6 1 | 64 |
Finding one term without expanding the rest
Exam questions often ask only for a particular term, and there is no need to build the whole expansion. The term containing bᵏ is C(n, k)·aⁿ⁻ᵏ·bᵏ, so the x²y³ term of (x + y)⁵ is C(5, 3)·x²·y³ = 10x²y³. Counting is the only trap: the term with bᵏ is the (k + 1)th term, because the first term has k = 0.
Two notes on scope. The signs for (a − b)ⁿ alternate because (−b)ᵏ is negative for odd k, so a difference raised to an even power ends positive and to an odd power ends negative. And this page handles non-negative integer n, where the expansion terminates; for a fractional or negative exponent the binomial series is infinite and converges only for |x| < 1, which is a different object covered on the series page.