Learn9709Paper 1 · Pure Mathematics 1SeriesBinomial expansion

Series · Subtopic 1 of 4

The Binomial Expansion

How to expand $(a+b)^n$ without multiplying out $n$ brackets, using Pascal's triangle or $\binom{n}{r}$ , and how to pick out one term, such as the coefficient of $x^3$ or the term independent of $x$ .

Cambridge 9709 Paper 1Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1AQA 7357 Paper 2≈ 45 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Use Pascal's triangle and $\binom{n}{r}=\dfrac{n!}{r!\,(n-r)!}$ .
Expand $(a+b)^n$ for a positive integer $n$ , in ascending or descending powers.
Find a particular coefficient, including the term independent of $x$ .
Find coefficients in products such as $(1+2x)(2-x)^5$ , and unknown constants from given coefficients.

Pause and think

Expand $(a+b)^2$ and $(a+b)^3$ . Write down just the numbers in front of each term. Do you recognise the pattern?

Section 2 of 14

Core idea

Start hereExpanding without the slog

$(a+b)^2$ is easy. $(a+b)^3$ is a bit of work. $(2+3x)^{10}$ would take an afternoon of multiplying brackets. Yet there's a pattern so regular that you can write down any term, say the $x^7$ term, directly. The pattern lives in Pascal's triangle, and it comes from counting choices.

The big idea

$(a+b)^n=\displaystyle\sum_{r=0}^{n}\binom nra^{n-r}b^r$ for positive whole $n$ . The coefficients $\binom nr=\dfrac{n!}{r!(n-r)!}$ are the numbers in Pascal's triangle and count the ways to choose $r$ $b$ 's from $n$ brackets.

1. The pattern

\begin{aligned}(a+b)^1&=a+b\\(a+b)^2&=a^2+2ab+b^2\\(a+b)^3&=a^3+3a^2b+3ab^2+b^3\end{aligned}

Pascal's triangle

Row

n

(counting the top as row

0

) gives the coefficients of

(a+b)^n

. Row

4

1,4,6,4,1

Powers of $a$ go down from $n$ ; powers of $b$ go up from $0$ ; each term's powers add to $n$ .
Coefficients: a row of Pascal's triangle (each number is the sum of the two above).

2. Why the coefficients count choices

The coefficient of

a^2b

(a+b)^3

$(a+b)^3=(a+b)(a+b)(a+b)$
Three brackets.
$\text{each term picks }a\text{ or }b\text{ from each bracket}$
That's how multiplying out works.
$a^2b\text{ arises from: }baa,\ aba,\ aab$
Choose which one of the three brackets supplies the $b$ .
$\text{coefficient}=3=\binom31$
The number of ways to choose $1$ bracket from $3$ .

In general the coefficient of

a^{n-r}b^r

\binom nr

, the number of ways to choose

r

brackets from

n

Binomial theorem

(a+b)^n=a^n+\binom n1a^{n-1}b+\binom n2a^{n-2}b^2+\cdots+b^n

$n!$: " $n$ factorial" $=n(n-1)(n-2)\cdots1$ ; $0!=1$ .
$\binom nr$: $\dfrac{n!}{r!(n-r)!}$ , also written $\,{}^nC_r$ .

3. Expanding and finding terms

Worked exampleExpand $(2+3x)^4$

Expand fully.

Goal: Five simplified terms.
Strategy: Row 4 of Pascal: $1,4,6,4,1$ ; bracket the whole of each term before powering.

$2^4+4(2^3)(3x)+6(2^2)(3x)^2+4(2)(3x)^3+(3x)^4$
$a=2$ , $b=3x$ .
$=16+96x+216x^2+216x^3+81x^4$
$(3x)^2=9x^2$ , $(3x)^3=27x^3$ , $(3x)^4=81x^4$ .

Answer

16+96x+216x^2+216x^3+81x^4

✓ CheckPut

x=1

(2+3)^4=625

and

16+96+216+216+81=625

✓.

Worked exampleTerm independent of $x$ in $\left(x^2+\dfrac2x\right)^6$

Find the constant term.

Goal: The term with $x^0$ .
Strategy: Write the general term, simplify its power of $x$ , set the power to $0$ .

$\binom6r(x^2)^{6-r}\left(\tfrac2x\right)^r$
General term.
$=\binom6r2^rx^{12-2r}x^{-r}=\binom6r2^rx^{12-3r}$
Index laws.
$12-3r=0\Rightarrow r=4$
Constant term.
$\binom64\cdot2^4=15\times16=240$
Evaluate.

Answer

240

✓ Check

\binom64=\frac{6\cdot5}{2}=15

✓.

Common misconceptionNot bracketing

(3x)^2=9x^2

, not

3x^2

. With

(1-2x)^n

take

b=-2x

, so signs alternate.

Pause and think

Use the first three terms of $(1+0.01)^8$ to estimate $1.01^8$ .

Summary

Coefficients $\binom nr$ count choices.
General term $\binom nra^{n-r}b^r$ .
Bracket whole terms; check with $x=1$ .

Binomial: Two-term expression.
Factorial: $n!=n(n-1)\cdots1$ .

HistoryPascal was not first. The triangle is named after Blaise Pascal, who wrote about it in 1654. But it appears in the work of the Persian poet-mathematician Omar Khayyam around 1100, in a 1261 book by the Chinese mathematician Yang Hui (in China it is still called Yang Hui's triangle), and its patterns were studied in India even earlier.

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