The binomial expansion lets you expand expressions of the form (a + bx)ⁿ without multiplying everything out by hand. At A-Level it splits into two distinct cases that behave completely differently, and a large share of the marks lost on this topic come from applying the wrong case, or the right case with the wrong validity condition.
Case 1: n is a positive integer
When n is a positive whole number, (a + bx)ⁿ expands into exactly n + 1 terms, and the expansion is exact — it is not an approximation and there is no restriction on x.
The general term uses binomial coefficients, written ⁿCᵣ or (n choose r), which count the number of ways to choose r items from n. The expansion is:
Worked example
Expand (2 + 3x)⁴ in ascending powers of x.
| r | Coefficient ⁴Cᵣ | Term |
|---|---|---|
| 0 | 1 | 2⁴ = 16 |
| 1 | 4 | 4 × 2³ × 3x = 96x |
| 2 | 6 | 6 × 2² × (3x)² = 216x² |
| 3 | 4 | 4 × 2 × (3x)³ = 216x³ |
| 4 | 1 | (3x)⁴ = 81x⁴ |
So (2 + 3x)⁴ = 16 + 96x + 216x² + 216x³ + 81x⁴. A very common slip here is forgetting to raise the 2 to the correct power in each term, or forgetting to cube or raise the 3x fully — always raise both parts of each term to the power shown in the table, not just the x part.
Finding a single term without the full expansion
Exam questions often ask only for the coefficient of x³, for example, rather than the whole expansion. Use the general term formula directly: the term containing xʳ in (a + bx)ⁿ is ⁿCᵣ aⁿ⁻ʳ (bx)ʳ. Substitute the value of r that matches the power you need, and you avoid expanding terms you don't need.
Case 2: n is negative or a fraction
When n is not a positive integer — it might be negative, or a fraction like ½ or −2 — the expansion no longer terminates. It becomes an infinite series, and it is only a valid approximation for certain values of x. This is where the topic changes character completely.
Two things change compared with Case 1: the series never ends (you are only ever asked for the first few terms), and there is now a restriction on x for the expansion to be valid at all. Both of these must be stated in a full answer — a correct expansion with no validity statement typically loses a mark.
Worked example: expression not already in the form (1 + x)ⁿ
Expand (2 + x)⁻¹ up to the term in x², stating the range of values of x for which the expansion is valid.
- Factor out the 2 so the bracket starts with 1: (2 + x)⁻¹ = 2⁻¹(1 + x/2)⁻¹ = ½(1 + x/2)⁻¹.
- Apply the formula with n = −1 and x replaced by x/2: (1 + x/2)⁻¹ = 1 + (−1)(x/2) + [(−1)(−2)/2!](x/2)² + ... = 1 − x/2 + x²/4 − ...
- Multiply by the ½ factored out at the start: ½ − x/4 + x²/8 − ...
- State validity: the expansion of (1 + x/2)⁻¹ is valid for |x/2| < 1, i.e. |x| < 2.
Why the validity condition changes
The validity condition always refers to the bracket in the form (1 + u), where |u| < 1. If you rewrite (2 + x)⁻¹ as ½(1 + x/2)⁻¹, the condition is on x/2, not on x directly — which is why the final range becomes |x| < 2 rather than |x| < 1. Always convert the condition back to plain x at the end, and don't leave it in terms of the substituted variable.
Using expansions to approximate numbers
A frequent application is using a binomial expansion to approximate a surd or a value like 1.02^(−3) to a number of decimal places, by choosing a small value of x that makes the expression match a known number. This tests whether you understand that the expansion is only accurate near x = 0 — the smaller x is, the better the approximation, which is exactly why questions choose a small value of x deliberately.
Common errors summary
| Error | Consequence |
|---|---|
| Using ⁿCᵣ notation/factorial formula when n is not a positive integer | Formula does not apply — coefficients will be wrong |
| Not factoring out the leading constant before expanding | Wrong coefficients and wrong validity range |
| Forgetting the validity condition altogether | Lost method/accuracy marks even with correct coefficients |
| Sign errors in negative n terms | Alternating signs are easy to drop, especially in the x³ term |
| Not squaring/cubing both parts of a bracket in Case 1 | Coefficients too small by a power of the constant |
How the two cases compare
| n positive integer | n negative or fractional | |
|---|---|---|
| Number of terms | Finite: n + 1 terms | Infinite series |
| Exact or approximate | Exact | Approximate, valid only for certain x |
| Coefficients | ⁿCᵣ (binomial coefficients) | n(n − 1)(n − 2).../r! pattern |
| Validity restriction | None | |x| < 1 (or scaled version) |
Once you separate these two cases clearly in your mind, most of the confusion in this topic disappears — the mistake to avoid is treating them as one formula that always behaves the same way. Practise recognising which case a question is in before you write a single term, since that decision determines everything that follows.
Worked example combining both cases
Exam questions increasingly ask you to use a Case 1 expansion to help simplify a Case 2 one, or to combine two expansions before comparing coefficients. Suppose you are asked to find the first three terms of (1 + x)(1 − 2x)⁻² in ascending powers of x.
- Expand (1 − 2x)⁻² using the Case 2 formula with n = −2 and x replaced by −2x: 1 + (−2)(−2x) + [(−2)(−3)/2!](−2x)² + ... = 1 + 4x + 12x² + ...
- Multiply this by (1 + x), keeping only terms up to x²: (1 + x)(1 + 4x + 12x²) = 1 + 4x + 12x² + x + 4x² + ... = 1 + 5x + 16x² + ...
- State validity: the expansion of (1 − 2x)⁻² requires |−2x| < 1, i.e. |x| < ½.
Comparing coefficients using a binomial expansion
A related question style gives you an expansion with unknown constants and asks you to find them by comparing coefficients of matching powers of x on both sides of an identity. This tests exactly the same expansion skills, but in reverse — you expand one side, then match term by term against the given expansion on the other side.
Worked example
Given that (1 + ax)⁵ = 1 + 20x + bx² + ... for constants a and b, find a and b.
- Expand (1 + ax)⁵ using Case 1 with n = 5: 1 + 5(ax) + 10(ax)² + ... = 1 + 5ax + 10a²x² + ...
- Compare the coefficient of x: 5a = 20, so a = 4.
- Compare the coefficient of x²: b = 10a² = 10(16) = 160.
Frequently confused points
Why does ⁿCᵣ not work when n is fractional or negative?
ⁿCᵣ = n! / (r!(n − r)!) relies on factorials, which are only defined for non-negative integers. When n is negative or fractional, (n − r)! is meaningless, so the formula breaks down entirely — it is not simply less accurate, it cannot be evaluated at all. This is why Case 2 uses the alternative pattern n(n − 1)(n − 2).../r! instead, which is defined for any rational n.
Does the validity condition ever change the answer itself?
No — the coefficients of the expansion are fixed by the formula regardless of x. The validity condition only tells you the range of x for which the infinite series actually converges to the correct value. Outside that range, the terms of the series do not settle down and the expansion becomes useless as an approximation, even though you can still write the algebraic terms down.
A revision checklist
- Identify whether n is a positive integer (Case 1, exact, finite) or not (Case 2, approximate, infinite) before writing a single term.
- In Case 2, always factor out the leading constant so the bracket reads exactly (1 + something).
- State the validity condition in terms of plain x, not the substituted expression.
- When combining two expansions, expand each fully to the required power first, then multiply.
- Double-check every negative sign when n is negative — the pattern n(n − 1)(n − 2)... generates alternating signs that are easy to lose.