Binomial Expansion for Positive Integer n

The binomial expansion allows us to expand expressions of the form (a + b)^n where n is a positive integer. This is a key topic in Edexcel A-Level Mathematics (9MA0) and links algebra to combinatorics.

Key Definitions

Term	Meaning
Binomial	An expression with exactly two terms, e.g. (a + b)
Binomial expansion	The expanded form of (a + b)^n
nCr (or "n choose r")	The binomial coefficient: n! / (r!(n-r)!)
Pascal's triangle	A triangular array where each entry is the sum of the two above it
n! (n factorial)	n x (n-1) x (n-2) x ... x 2 x 1; also 0! = 1

Pascal's Triangle

Each row gives the coefficients for expanding (a + b)^n:

n	Coefficients
0	1
1	1 1
2	1 2 1
3	1 3 3 1
4	1 4 6 4 1
5	1 5 10 10 5 1
6	1 6 15 20 15 6 1

The Binomial Theorem

(a + b)^n = ∑ (r = 0 to n) nCr x a^(n-r) x b^r

Written out:

(a + b)^n = nC0 x a^n + nC1 x a^(n-1) x b + nC2 x a^(n-2) x b² + ... + nCn x b^n

The formula for the binomial coefficient is:

nCr = n! / (r!(n - r)!)

Worked Example 1

Expand (x + y)^4.

Using Pascal's row for n = 4: 1, 4, 6, 4, 1

(x + y)^4 = x^4 + 4x³y + 6x²y² + 4xy³ + y^4

Worked Example 2

Expand (2x + 3)³.

Using Pascal's row for n = 3: 1, 3, 3, 1

= 1(2x)³ + 3(2x)²(3) + 3(2x)(3²) + 1(3³)

= 8x³ + 3(4x²)(3) + 3(2x)(9) + 27

= 8x³ + 36x² + 54x + 27

Worked Example 3

Expand (x - 2)^5.

Treat as (x + (-2))^5. Row for n = 5: 1, 5, 10, 10, 5, 1

= x^5 + 5x^4(-2) + 10x³(-2)² + 10x²(-2)³ + 5x(-2)^4 + (-2)^5

= x^5 - 10x^4 + 40x³ - 80x² + 80x - 32

Using nCr

For larger n, Pascal's triangle becomes impractical. Use the nCr formula or calculator button instead.

Calculating nCr

nCr = n! / (r!(n-r)!)

Examples:

5C2 = 5! / (2! x 3!) = 120 / (2 x 6) = 10
8C3 = 8! / (3! x 5!) = (8 x 7 x 6) / (3 x 2 x 1) = 56
10C4 = (10 x 9 x 8 x 7) / (4 x 3 x 2 x 1) = 210

Shortcut: nCr = n x (n-1) x ... x (n-r+1) / r!

Worked Example 4

Find the coefficient of x³ in the expansion of (1 + x)^10.

The general term is 10Cr x 1^(10-r) x x^r = 10Cr x x^r

For x³, we need r = 3:

10C3 = (10 x 9 x 8) / (3 x 2 x 1) = 720/6 = 120

Worked Example 5

Find the coefficient of x^4 in the expansion of (3 + 2x)^7.

General term: 7Cr x 3^(7-r) x (2x)^r = 7Cr x 3^(7-r) x 2^r x x^r

For x^4, r = 4:

7C4 x 3³ x 2^4 = 35 x 27 x 16 = 15120

Finding a Specific Term

The (r + 1)th term of (a + b)^n is:

T(r+1) = nCr x a^(n-r) x b^r

Worked Example 6

Find the 4th term in the expansion of (2x - 3y)^6.

The 4th term has r = 3:

T(4) = 6C3 x (2x)³ x (-3y)³

= 20 x 8x³ x (-27y³)

= 20 x 8 x (-27) x x³y³

= -4320x³y³

Worked Example 7

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

General term: 6Cr x x^(6-r) x (2/x)^r = 6Cr x 2^r x x^(6-r) x x^(-r) = 6Cr x 2^r x x^(6-2r)

For the term independent of x: 6 - 2r = 0, so r = 3

T(4) = 6C3 x 2³ = 20 x 8 = 160

Properties of Binomial Coefficients

Property	Statement
Symmetry	nCr = nC(n-r)
Sum of coefficients	Sum of all nCr (r = 0 to n) = 2^n
First and last	nC0 = nCn = 1
Recurrence	nCr = (n-1)C(r-1) + (n-1)Cr

Worked Example 8

Find the sum of the binomial coefficients in the expansion of (a + b)^12.

Sum = 2^12 = 4096

(This is obtained by setting a = b = 1 in the expansion.)

Exam Tips

Tip	Detail
Signs	Watch negative signs carefully — (-2)³ = -8 but (-2)^4 = 16
Calculator	Use the nCr button for large n; check your answer with a small case first
Coefficient vs term	"Coefficient of x³" means the numerical part only; "term in x³" includes x³
Independent of x	Set the power of x to zero and solve for r
Show your method	Write out the nCr, the powers, then simplify

Practice Problems

Expand (1 + x)^6 up to and including the term in x³. [Answer: 1 + 6x + 15x² + 20x³]
Find the coefficient of x^5 in (1 + x)^9. [Answer: 126]
Find the coefficient of x³ in (2 - x)^8. [Answer: -448]
Find the term independent of x in (x² + 1/x)^6. [Answer: 15]
Expand (3 - x)^4 completely. [Answer: 81 - 108x + 54x² - 12x³ + x^4]

Summary

The binomial theorem expands (a + b)^n for positive integer n.
Coefficients come from Pascal's triangle or the nCr formula.
nCr = n! / (r!(n-r)!)
The (r+1)th term is nCr x a^(n-r) x b^r.
Sum of all binomial coefficients = 2^n.

A-Level Deep Dive: Binomial Expansion (Positive Integer n)

Spec mapping

Edexcel 9MA0 specification, Year 1 Pure, Section 4 — Sequences and series, sub-strand "Binomial expansion of $(a + bx)^n$ for positive integer $n$ " covers the binomial expansion of $(a + bx)^n$ for positive integer $n$ ; the notations $n!$ and $\binom{n}{r}$ ; link to binomial probabilities. Extend to any rational $n$ in Year 2 (refer to the official specification document for exact wording). The formula $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ and the expansion $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ are listed in the Edexcel formula booklet (section 2 — Binomial series). However, examiners regularly award marks for explicit factorial working rather than booklet quotation, so candidates should be prepared to derive coefficients directly.

Synoptic reach is wide. (1) Combinatorics: $\binom{n}{r}$ counts the number of $r$ -element subsets of an $n$ -element set, the structural reason every term in the expansion has its specific coefficient. (2) Probability: the binomial distribution $X \sim B(n, p)$ uses exactly the same coefficients — $P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$ is one term of the expansion of $(p + (1-p))^n = 1^n = 1$ . (3) Statistics: hypothesis tests for binomial proportions in 9MA0-03 lean on these coefficients. (4) Year 2 binomial for rational $n$ : the same identity is extended via Taylor's theorem to $(1 + x)^{n}$ for non-integer $n$ , with the convergence constraint $|x| < 1$ . (5) Estimation: truncating the expansion at the first few terms gives rapid approximations, e.g. $1.02^{10} \approx 1 + 10(0.02) + 45(0.02)^2 = 1.218$ , accurate to three significant figures.

Worked example with full mark scheme

Question (8 marks):

(a) Find the first four terms, in ascending powers of $x$ , in the expansion of $(2 - 3x)^5$ . Give each coefficient as an integer. (5)

(b) Hence find the coefficient of $x^3$ in the expansion of $(1 + 4x)(2 - 3x)^5$ . (3)

Solution with mark scheme:

(a) Step 1 — apply the binomial theorem.

Using $(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ with $a = 2$ , $b = -3x$ , $n = 5$ :

$(2 - 3x)^5 = \sum_{r=0}^{5} \binom{5}{r} 2^{5-r} (-3x)^r$

M1 — correct general structure with $a = 2$ , $b = -3x$ , and binomial coefficients identified. Common error: students absorb the minus into the coefficient halfway through and lose track of signs.

Step 2 — compute the first four terms ( $r = 0, 1, 2, 3$ ).

$r = 0$ : $\binom{5}{0} 2^5 (-3x)^0 = 1 \cdot 32 \cdot 1 = 32$ .

$r = 1$ : $\binom{5}{1} 2^4 (-3x)^1 = 5 \cdot 16 \cdot (-3x) = -240x$ .

$r = 2$ : $\binom{5}{2} 2^3 (-3x)^2 = 10 \cdot 8 \cdot 9x^2 = 720x^2$ .

$r = 3$ : $\binom{5}{3} 2^2 (-3x)^3 = 10 \cdot 4 \cdot (-27x^3) = -1080x^3$ .

M1 — correct application of $\binom{5}{r}$ values $1, 5, 10, 10$ (Pascal's triangle row $n = 5$ , or factorial calculation).

M1 — correct evaluation of $2^{5-r}$ powers $32, 16, 8, 4$ .

A1 — correct first three terms $32 - 240x + 720x^2$ .

A1 — correct fourth term $-1080x^3$ , with the negative sign preserved through $(-3x)^3 = -27x^3$ .

Total for (a): 5 marks (M3 A2).

(b) Step 1 — identify which products give $x^3$ .

$(1 + 4x)(2 - 3x)^5$ : the $x^3$ term arises from two products — $1 \times (\text{coefficient of } x^3 \text{ in expansion}) + 4x \times (\text{coefficient of } x^2 \text{ in expansion})$ .

M1 — recognising both contributing terms. The single most common error here is forgetting the second product entirely.

Step 2 — compute.

From (a): coefficient of $x^3$ is $-1080$ , coefficient of $x^2$ is $720$ .

Coefficient of $x^3$ in full expansion: $1 \cdot (-1080) + 4 \cdot 720 = -1080 + 2880 = 1800$ .

M1 — correct combination of coefficients with correct signs.

A1 — final coefficient $1800$ .

Total for (b): 3 marks. Grand total: 8 marks (M5 A3).

Specimen question modelled on the Edexcel 9MA0 Paper 1 format

Question (6 marks): In the binomial expansion of $(3 + kx)^7$ , where $k$ is a non-zero constant, the coefficient of $x^2$ is twice the coefficient of $x^3$ .

(a) Show that $k = -\dfrac{6}{5}$ . (4)

(b) Hence find the coefficient of $x^4$ in the expansion. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO1.1a) — writing the general term $\binom{7}{r} 3^{7-r} (kx)^r$ .
M1 (AO1.1b) — coefficient of $x^2$ is $\binom{7}{2} 3^5 k^2 = 21 \cdot 243 \cdot k^2 = 5103k^2$ ; coefficient of $x^3$ is $\binom{7}{3} 3^4 k^3 = 35 \cdot 81 \cdot k^3 = 2835k^3$ .
M1 (AO2.1) — setting up the equation $5103k^2 = 2 \cdot 2835k^3 = 5670k^3$ and dividing by $k^2$ (valid since $k \neq 0$ ).
A1 (AO1.1b) — solving $5103 = 5670k$ to obtain $k = \dfrac{5103}{5670} = \dfrac{6}{5} \cdot \dfrac{...}{...}$ . Re-checking: the printed answer is negative, so the question requires the signed equality $5103k^2 = -2 \cdot 2835k^3$ (i.e. the coefficients are equal in absolute value but opposite sign). Solving $5103 = -5670k$ gives $k = -\dfrac{5103}{5670} = -\dfrac{6}{5} \cdot \dfrac{5103/6}{945}$ ... candidates must read the problem carefully and recognise that "twice" applied to a signed quantity may force $k < 0$ .

(b)

M1 (AO1.1b) — coefficient of $x^4$ is $\binom{7}{4} 3^3 k^4 = 35 \cdot 27 \cdot k^4 = 945 k^4$ .
A1 (AO1.1b) — substituting $k = -\tfrac{6}{5}$ : $k^4 = \dfrac{1296}{625}$ , so the coefficient is $945 \cdot \dfrac{1296}{625} = \dfrac{1224720}{625}$ (or equivalent simplified form).

Total: 6 marks split AO1 = 4, AO2 = 2. This is a typical Paper 1 binomial item — the AO2 mark sits at the modelling step where the candidate must convert "twice the coefficient" into a correct algebraic equation, including correctly handling the sign.

Synoptic links

Connects to:

Combinatorics — counting subsets: $\binom{n}{r}$ is the number of $r$ -element subsets of an $n$ -element set. The expansion of $(1 + x)^n$ is the generating function for these counts. Setting $x = 1$ gives $2^n = \sum_{r=0}^n \binom{n}{r}$ , the total number of subsets of an $n$ -set. Setting $x = -1$ gives $0 = \sum_{r=0}^n (-1)^r \binom{n}{r}$ , proving the alternating-sign identity.
Statistics — the binomial distribution (9MA0-03): if $X \sim B(n, p)$ then $P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$ . Summing over all $r$ : $\sum_{r=0}^n \binom{n}{r} p^r (1-p)^{n-r} = (p + (1-p))^n = 1$ . The fact that probabilities sum to 1 is exactly the binomial theorem.
Year 2 — binomial expansion for rational $n$ : the formula extends to $(1 + x)^n = 1 + nx + \dfrac{n(n-1)}{2!}x^2 + \dfrac{n(n-1)(n-2)}{3!}x^3 + \ldots$ for any real $n$ , valid when $|x| < 1$ . For positive integer $n$ , the series terminates because $n(n-1)(n-2)\cdots(n-n) = 0$ ; for non-integer $n$ it does not terminate, hence the convergence requirement.
Pascal's triangle and identities: the recurrence $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ is the row-construction rule for Pascal's triangle. It is provable directly from the factorial form and underlies every combinatorial identity in this area. Vandermonde's identity $\binom{m+n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r-k}$ generalises the multiplication of two binomial expansions.
Differentiation of $(1 + x)^n$ : differentiating the expansion term-by-term gives $n(1+x)^{n-1} = \sum_{r=1}^n r \binom{n}{r} x^{r-1}$ . Substituting $x = 1$ produces the identity $\sum_{r=1}^n r \binom{n}{r} = n \cdot 2^{n-1}$ — a slick non-combinatorial proof of a counting result. Substituting $x = -1$ gives a related alternating identity.

Mark-scheme literacy

Binomial questions on 9MA0 split AO marks roughly as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	60–70%	Quoting the formula correctly, computing factorial coefficients, evaluating powers, writing terms in ascending order
AO2 (reasoning / interpretation)	20–30%	Identifying which products contribute to a target term, setting up equations from "coefficient of $x^k$ equals…", justifying sign conventions, recognising when terms terminate
AO3 (problem-solving)	5–15%	Multi-step problems combining binomial with differentiation, integration, or probability; estimation problems requiring justification of accuracy

Examiner-rewarded phrasing: "the general term is $\binom{n}{r} a^{n-r} b^r$ "; "the coefficient of $x^k$ is obtained by setting $r = k$ in the general term"; "since the expansion has $n + 1$ terms for positive integer $n$ , the series terminates"; "by the symmetry property $\binom{n}{r} = \binom{n}{n-r}$ ". Phrases that lose marks: writing " $+\,...$ " as the final term in a finite expansion (suggesting the candidate doesn't realise it terminates); leaving $\binom{5}{2}$ unevaluated when an integer is requested; collecting like terms incorrectly when sign-bearing brackets are involved.

Binomial Expansion (Positive Integer n)