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MathematicsAlgebra

The binomial expansion formula

The binomial expansion formula expands a bracket raised to a power, (a + b)ⁿ, without multiplying it out by hand. For a positive whole number n it is (a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + bⁿ, where the coefficients ⁿCᵣ come from Pascal's triangle or the nCr button on your calculator. At A-level, a second version, (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + …, works for negative and fractional n too, as long as |x| < 1.

The binomial expansion formula for (a + b)ⁿ

(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^nFor n a positive whole number

Each term follows the same pattern. The power of a starts at n and goes down by one each time. The power of b starts at 0 and goes up by one. The two powers in every term add up to n. There are n + 1 terms in total.

The general term, the term containing bʳ, is:

\binom{n}{r}a^{n-r}b^r

The binomial coefficients, ⁿCᵣ

The numbers in front, written ⁿCᵣ or (n r) in brackets, are called binomial coefficients and are read "n choose r":

\binom{n}{r} = \frac{n!}{r!(n - r)!}

where n! means n × (n − 1) × … × 2 × 1. So ⁵C₂ = 5! ÷ (2! × 3!) = 120 ÷ 12 = 10. On a calculator, type 5, then nCr, then 2.

Pascal's triangle

For small powers, the coefficients are quicker to read from Pascal's triangle. Each row starts and ends with 1, and every other number is the sum of the two above it. Row n gives the coefficients for (a + b)ⁿ.

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
n = 6:    1 6 15 20 15 6 1

So (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.

Why the coefficients are ⁿCᵣ

Write (a + b)⁴ as four brackets multiplied together: (a + b)(a + b)(a + b)(a + b). Expanding means picking either a or b from each bracket and multiplying, in every possible way. To get a²b², you pick b from exactly 2 of the 4 brackets. The number of ways to choose 2 brackets out of 4 is ⁴C₂ = 6, which is why the a²b² term is 6a²b². The same counting gives every coefficient, and it is the same ⁿCᵣ that appears in the binomial distribution in statistics, which is where that distribution gets its name.

How to do a binomial expansion

  1. Identify a, b and n. Keep any minus sign with b, and any number with x: in (3 − 2x)⁵, a = 3, b = −2x and n = 5.
  2. Write the coefficients from Pascal's triangle or nCr.
  3. Write the powers: a down from n, b up from 0.
  4. Work out each term, putting brackets round b so its number and sign are raised to the power too.
  5. Simplify, and stop at the number of terms the question asks for.

Example 1: a full expansion

Expand (x + 2)⁴.

Here a = x, b = 2 and n = 4, so the coefficients are 1, 4, 6, 4, 1.

(x + 2)^4 = x^4 + 4x^3(2) + 6x^2(2)^2 + 4x(2)^3 + (2)^4
= x^4 + 8x^3 + 24x^2 + 32x + 16

Example 2: the first terms in ascending powers

Find the first four terms of (3 − 2x)⁵ in ascending powers of x.

Ascending powers means start with the term with no x. Here a = 3, b = −2x and n = 5, so the coefficients are 1, 5, 10, 10.

3^5 + 5(3^4)(-2x) + 10(3^3)(-2x)^2 + 10(3^2)(-2x)^3
= 243 - 810x + 1080x^2 - 720x^3

The signs alternate because b is negative: odd powers of −2x are negative.

Example 3: finding one coefficient

Find the coefficient of x³ in the expansion of (3 + 2x)⁶.

You do not need the whole expansion. The x³ term has r = 3, so use the general term:

\binom{6}{3}(3)^3(2x)^3 = 20 \times 27 \times 8x^3 = 4320x^3

The coefficient is 4320. The most common slip is forgetting to cube the 2 as well as the x.

Example 4: the term independent of x

Find the term independent of x in (x + 2/x)⁶.

The general term is ⁶Cᵣ x⁶⁻ʳ (2/x)ʳ = ⁶Cᵣ 2ʳ x⁶⁻²ʳ. Independent of x means the power of x is 0, so 6 − 2r = 0 and r = 3.

\binom{6}{3} \times 2^3 = 20 \times 8 = 160

The term independent of x is 160.

Example 5: an estimate

Use the first three terms of (1 + x)⁸ to estimate 1.02⁸.

(1 + x)⁸ = 1 + 8x + 28x² + …. With x = 0.02:

1 + 8(0.02) + 28(0.02)^2 = 1 + 0.16 + 0.0112 = 1.1712

The true value is 1.17166, so three terms are already accurate to 3 decimal places. The later terms contain x³, x⁴ and so on, which are tiny when x is small.

Try it

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The general binomial expansion: year 2 A-level

For any rational n, including negative numbers and fractions, the expansion of (1 + x)ⁿ never stops: it becomes an infinite series. It only works, or converges, when |x| < 1.

(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \dotsValid for |x| < 1 when n is negative or a fraction

When n is a positive whole number, the coefficients eventually hit zero and the series stops, giving the same answer as the first formula. For other values of n, you write out as many terms as the question asks for.

Example 6: a negative power

Find the first three terms of (1 − 2x)⁻³, and state when the expansion is valid.

Here n = −3, and the x in the formula is replaced by −2x.

1 + (-3)(-2x) + \frac{(-3)(-4)}{2}(-2x)^2 = 1 + 6x + 24x^2

The expansion is valid when |−2x| < 1, which means |x| < ½.

Example 7: a fractional power, and an approximation

Expand √(1 + x) up to the x² term, and use it to estimate √1.1.

√(1 + x) = (1 + x)^(1/2), so n = ½.

1 + \frac{1}{2}x + \frac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2}x^2 = 1 + \frac{1}{2}x - \frac{1}{8}x^2

With x = 0.1: 1 + 0.05 − 0.00125 = 1.04875. The true value of √1.1 is 1.04881, so the estimate is out by less than 0.0001.

The graph shows why the estimate is good for small x and bad for large x. Here x is the input and y is the error: the true √(1 + x) minus the three-term estimate.

y = \sqrt{1+ x}-\left(1+\frac{ x}{2}-\frac{{ x}^{2}}{8}\right)
-1-0.8-0.6-0.4-0.20.20.40.60.811.21.41.61.82-0.4-0.20.2x = 0.1: error tinyx = 1: error 0.04
Drag to pan, scroll to zoom. x from -1 to 2, y from -0.496 to 0.353.

Close to x = 0 the error is almost zero. As x moves towards ±1 it grows, and beyond the range |x| < 1 the infinite series stops converging altogether.

Expanding (a + bx)ⁿ: take out a factor first

The general formula needs the bracket to start with 1. If it starts with another number, take that number out as a factor, raised to the power n.

(4 - x)^{-\frac{1}{2}} = 4^{-\frac{1}{2}}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}} = \frac{1}{2}\left(1 - \frac{x}{4}\right)^{-\frac{1}{2}}

Then expand with n = −½ and −x/4 in place of x:

\frac{1}{2}\left(1 + \frac{x}{8} + \frac{3x^2}{128} + \dots\right) = \frac{1}{2} + \frac{x}{16} + \frac{3x^2}{256} + \dots

This is valid when |x/4| < 1, which means |x| < 4. In general, (a + bx)ⁿ is valid for |x| < |a/b|.

Here is a quick check before the mistakes.

practice
Pick an answer, then check it.
For which values of x is the expansion of (1 + 3x)^(1/2) valid?

Common mistakes

  • Forgetting to raise the number with x. In (3 + 2x)⁶, the term with r = 3 contains (2x)³ = 8x³, not 2x³.
  • Losing a negative sign. Keep the minus inside the bracket: (−2x)² = 4x² but (−2x)³ = −8x³.
  • Not taking out the factor properly. (4 − x)^(−1/2) becomes 4^(−1/2) × (1 − x/4)^(−1/2), and 4^(−1/2) is ½, not 4 or −2.
  • Forgetting the validity condition. With a negative or fractional power, always state when the expansion is valid.
  • Using the infinite series when n is a positive whole number. Then the expansion is finite and exact.
  • Mixing up ascending and descending powers. Ascending powers of x means start with the constant term.

Is the binomial expansion formula in the formula booklet?

Yes, for A-level Maths. This was checked against the exam boards' booklets in November 2026.

ExamWhat the booklet gives
AS Maths (Edexcel)(a + b)ⁿ for n a positive whole number, with ⁿCᵣ = n!/(r!(n − r)!)
A-level Maths (Edexcel)(a + b)ⁿ, and (1 + x)ⁿ for |x| < 1 and rational n
AS and A-level Maths (AQA)(a + b)ⁿ with ⁿCᵣ, and (1 + x)ⁿ for |x| < 1 and rational n

So you do not need to memorise the formulas, but you do need to use them quickly and accurately: substituting the right a, b and n, handling signs and fractions, and stating validity.

Binomial expansion questions with answers

1. Expand (2 + x)⁴ fully.
Coefficients 1, 4, 6, 4, 1:
16 + 4(8)x + 6(4)x² + 4(2)x³ + x⁴ = 16 + 32x + 24x² + 8x³ + x⁴
2. Find the first three terms of (1 + 3x)⁷ in ascending powers of x.
1 + 7(3x) + 21(3x)² = 1 + 21x + 189x²
3. Find the coefficient of x² in (2 + 5x)⁴.
⁴C₂ × 2² × 5² = 6 × 4 × 25 = 600
4. Find the first three terms of (1 + x)⁻², and state when the expansion is valid.
1 + (−2)x + ((−2)(−3) ÷ 2)x² = 1 − 2x + 3x², valid for |x| < 1
5. Find the term independent of x in (x² + 1/x)⁶.
General term: ⁶Cᵣ (x²)⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x¹²⁻³ʳ.
12 − 3r = 0, so r = 4, and the term is ⁶C₄ = 15

Practise in a short timed paper

Binomial expansion is a regular in A-level pure maths papers. A-level Maths: Algebra and Functions gives you a short timed paper every day, with 8 questions worth 25 to 35 marks. In the US, the same practice opens as AP Precalculus. Check any expansion in the Binomial Expansion Calculator.

Paper resets in

Common questions

What is the binomial expansion formula?
(a + b)ⁿ = aⁿ + ⁿC₁aⁿ⁻¹b + ⁿC₂aⁿ⁻²b² + … + bⁿ, for n a positive whole number. The general term is ⁿCᵣaⁿ⁻ʳbʳ.
What is the binomial theorem?
The binomial theorem is the statement that the binomial expansion formula is true. People use the two names for the same thing.
When do you use Pascal's triangle instead of nCr?
Pascal's triangle is quickest for small powers, up to about 6. For larger powers, or when you need just one coefficient, use nCr on your calculator.
What is the general binomial expansion?
The year 2 A-level version: (1 + x)ⁿ = 1 + nx + n(n − 1)x²/2! + … for any rational n. It is an infinite series when n is negative or a fraction, and only valid for |x| < 1.
Why does the expansion need |x| &lt; 1?
For negative or fractional powers, the series goes on forever. If |x| < 1, the powers of x shrink and the terms add up to a fixed value. If |x| ≥ 1, they do not shrink, and the series has no sum.

Revision cards

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Maths · AlgebraBinomial expansion calculator: any power with stepsExpand (a + b)ⁿ with the binomial theorem and Pascal's triangle, find any term or coefficient, and expand negative and fractional powers as series.