What Is the Binomial Theorem?
The binomial theorem is a formula that tells you exactly how to expand an expression of the form (a + b)^n, where a and b are any numbers or variables and n is a non-negative integer.
Expanding (a + b)² by hand is easy — you get a² + 2ab + b². But what about (a + b)⁷ or (2x − 3)¹⁰? Multiplying out ten factors by hand is slow and error-prone. The binomial theorem gives you the full expansion directly, without doing the multiplication.
The pattern behind every expansion
The key observations are: each term has the form (coefficient) · a^(n−k) · b^k; the powers of a decrease from n down to 0, while the powers of b increase from 0 up to n; the sum of the two powers in every term is always n; and the expansion has exactly n + 1 terms.
The only piece that needs explaining is the coefficient on each term — and that is where Pascal's triangle and the nCr formula come in.
Pascal's Triangle and the Coefficients
Pascal's triangle is a triangular array of numbers that gives you the binomial coefficients directly. It is built using a single rule: each number is the sum of the two numbers directly above it, with 1's down both edges.
The first several rows look like this:
| Row | Entries |
|---|---|
| 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 of Pascal's triangle gives you the coefficients for the expansion of (a + b)^n, in order. For example, row 4 is 1, 4, 6, 4, 1, so (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
Pascal's triangle is perfect for small exponents (say, n ≤ 8) because you can build it quickly by hand. For larger exponents, writing out every row becomes tedious, and the general formula with nCr is more efficient.
But the triangle reveals a beautiful pattern: the coefficients are symmetric, and each row matches the row of combinations nC0, nC1, nC2, ..., nCn from counting theory.
The General Formula
n: (a + b)^n = Σ (from k=0 to n) C(n,k) · a^(n−k) · b^k. Written out, this is (a + b)^n = C(n,0)a^n + C(n,1)a^(n−1)b + C(n,2)a^(n−2)b² + ... + C(n,n)b^n.C(n,k) — also written nCk or "n choose k" — is the binomial coefficient, and it is calculated by C(n,k) = n! / (k! · (n−k)!).For example, C(5,2) = 5! / (2! · 3!) = 120 / (2 · 6) = 10, which matches the middle entry of row 5 in Pascal's triangle.
C(5,2) = 10Every binomial expansion follows the same three-part pattern in each term: a coefficient from Pascal's triangle, a power of a that starts at n and decreases, and a power of b that starts at 0 and increases. The two powers always sum to n, and the expansion has n + 1 terms numbered from k = 0 to k = n. Once you internalize this pattern, you can expand any binomial directly from the formula without hesitation.
Expanding (a+b)^n Step by Step
Let's expand (2x + 3)⁴ using the binomial theorem.
- Step 1 — Identify n
Here
n = 4, so the expansion will have5terms. - Step 2 — Get the coefficients from row 4 of Pascal's triangle
1, 4, 6, 4, 1. - Step 3 — Write out the structure
Let
a = 2xandb = 3. The expansion is:1·(2x)⁴·(3)⁰ + 4·(2x)³·(3)¹ + 6·(2x)²·(3)² + 4·(2x)¹·(3)³ + 1·(2x)⁰·(3)⁴. - Step 4 — Simplify each term
(2x)⁴ = 16x⁴, so the first term is16x⁴.4·(2x)³·3 = 4·8x³·3 = 96x³.6·(2x)²·9 = 6·4x²·9 = 216x².4·(2x)·27 = 8x·27 = 216x.1·1·81 = 81.
(2x + 3)⁴ = 16x⁴ + 96x³ + 216x² + 216x + 81The most common errors come from forgetting to raise the full factor (including any coefficient in front of x) to the power, and from sign errors when b is negative. For (a − b)^n, the signs alternate: +, −, +, −, ..., because (−b)^k is negative when k is odd.
Finding a Specific Term
Sometimes a problem doesn't ask for the full expansion — only a single term, like "find the term containing x⁵" or "find the 4th term." The binomial theorem lets you jump directly to any term without expanding everything before it.
k) is T_(k+1) = C(n,k) · a^(n−k) · b^k. Notice the indexing: k = 0 gives the 1st term, k = 1 gives the 2nd, and so on. So the (r)th term corresponds to k = r − 1.Find the 5th term in the expansion of (x + 2)¹⁰.
Here n = 10 and the 5th term means k = 4.
The term is C(10,4) · x^(10−4) · 2⁴ = 210 · x⁶ · 16 = 3360x⁶.
3360x⁶Find the term containing x³ in the expansion of (x + 1)⁷.
Here a = x, b = 1, n = 7, and we need a^(n−k) = x³, so n − k = 3, meaning k = 4.
The term is C(7,4)·x³·1⁴ = 35x³.
35x³This ability to grab one specific term is extremely useful on standardized tests and in problems where computing the full expansion would be overkill.
Practice Problems
Expand (x + 2)⁵.
Answer: Row 5 of Pascal's triangle is 1, 5, 10, 10, 5, 1, so (x + 2)⁵ = x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32.
x⁵ + 10x⁴ + 40x³ + 80x² + 80x + 32Expand (3y − 1)⁴.
Hint: Remember the alternating signs.
81y⁴ − 108y³ + 54y² − 12y + 1Find the 6th term in the expansion of (2x + y)⁸.
Hint: The 6th term corresponds to k = 5.
Answer: C(8,5) · (2x)³ · y⁵ = 56 · 8x³ · y⁵ = 448x³y⁵.
448x³y⁵Find the coefficient of x⁴ in the expansion of (x − 2)⁷.
Hint: Need x^(7−k) = x⁴, so k = 3.
The term is C(7,3) · x⁴ · (−2)³ = 35 · x⁴ · (−8) = −280x⁴. The coefficient is −280.
−280Use the binomial theorem to approximate (1.01)⁵.
Hint: Write it as (1 + 0.01)⁵ and use the first few terms.
Answer: ≈ 1 + 5(0.01) + 10(0.01)² + ... ≈ 1.0510.
≈ 1.0510If you want to verify an expansion or double-check a specific term before a test, scan your work with Solver AI to see every binomial coefficient, every power, and every simplification laid out step by step.