Binomial expansion

AS · P1 · 10 min

Multiplying out $(2 + x)^4$ by hand takes four rounds of brackets; $(2 + x)^{10}$ would take all afternoon. The binomial expansion writes down every term of $(a + b)^n$ directly, using the numbers $\binom{n}{r}$ , and lets you pick out a single term, such as the coefficient of $x^3$ or the term independent of $x$ , without expanding everything. In Paper 1 the power $n$ is always a positive integer, so the expansion has exactly $n + 1$ terms. A binomial question worth 3 to 5 marks appears on nearly every paper.

Where the coefficients come from

Expand $(a + b)^n$ for small $n$ and look at the coefficients:

\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 \\ (a + b)^4 &= a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{aligned}

Two patterns stand out. In each term the powers of $a$ and $b$ add up to $n$ , with the power of $a$ going down from $n$ to $0$ and the power of $b$ going up from $0$ to $n$ . And the coefficients form Pascal's triangle, where each number is the sum of the two above it:

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

Why: counting choices

$(a + b)^n$ is $n$ brackets multiplied together. Each term of the expansion comes from choosing either $a$ or $b$ from every bracket. The term $a^{n-r}b^r$ arises once for every way of choosing which $r$ brackets supply a $b$ . So its coefficient is the number of ways of choosing $r$ objects from $n$ , written $\binom{n}{r}$ .

Factorials and $\binom{n}{r}$

Definition

For a positive integer $n$ , $n$ factorial is $n! = n \times (n - 1) \times \cdots \times 2 \times 1$ , and $0! = 1$ .

The binomial coefficient is

\binom{n}{r} = {}^nC_r = \frac{n!}{r!\,(n - r)!} \qquad (0 \le r \le n)

For calculation, cancel the larger factorial:

\binom{n}{r} = \frac{n(n - 1)(n - 2)\cdots(n - r + 1)}{r!} \qquad\text{for example}\quad \binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

Useful facts: $\binom{n}{0} = \binom{n}{n} = 1$ , $\binom{n}{1} = n$ , $\binom{n}{2} = \dfrac{n(n - 1)}{2}$ , and the symmetry $\binom{n}{r} = \binom{n}{n - r}$ (choosing which $r$ to take is the same as choosing which $n - r$ to leave). Your calculator's ${}^nC_r$ key gives these directly.

The binomial theorem

Key result

For a positive integer $n$ ,

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

The general term, the $(r + 1)$ th term, is

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

In particular

(1 + x)^n = 1 + nx + \frac{n(n - 1)}{2!}x^2 + \frac{n(n - 1)(n - 2)}{3!}x^3 + \cdots + x^n

Both forms are in the list of formulae (MF19).

The term containing $b^r$ is the $(r + 1)$ th term, not the $r$ th, because the first term has $b^0$ .

Expanding (a + b) to the power n

Identify $a$ and $b$ , including signs and coefficients. For $\left(3 - \tfrac{x}{2}\right)^6$ , $a = 3$ and $b = -\tfrac{x}{2}$ .
Write each term as $\binom{n}{r}(a)^{n-r}(b)^r$ , with $a$ and $b$ in brackets.
Evaluate each part separately, then multiply.
Simplify each coefficient fully and write the terms in the order asked ("ascending powers of $x$ " means starting from the constant).

Finding one particular term

Write the general term $\binom{n}{r}(a)^{n-r}(b)^r$ , with $a$ and $b$ as expressions in $x$ .
Collect the power of $x$ in terms of $r$ .
Set that power equal to the one you want (for example $0$ for the term independent of $x$ ) and solve for $r$ . It must be a whole number from $0$ to $n$ .
Substitute $r$ back to find the term or its coefficient.

Products of two brackets

To find a coefficient in something like $(1 + 3x)(2 - x)^5$ , expand the binomial only as far as you need, then collect every pair of terms whose powers of $x$ add up to the one you want.

Worked examples

A full expansion

Expand $(2 + x)^4$ fully.

Solution

Coefficients $1, 4, 6, 4, 1$ , with $a = 2$ , $b = x$ :

\begin{aligned} (2 + x)^4 &= 2^4 + 4(2^3)x + 6(2^2)x^2 + 4(2)x^3 + x^4 \\ &= 16 + 32x + 24x^2 + 8x^3 + x^4 \end{aligned}

Negative and fractional terms

Find the first three terms, in ascending powers of $x$ , of $\left(3 - \dfrac{x}{2}\right)^6$ .

Solution

$a = 3$ and $b = -\dfrac{x}{2}$ . Keep the sign inside the bracket.

\begin{aligned} &3^6 + \binom{6}{1}3^5\left(-\frac{x}{2}\right) + \binom{6}{2}3^4\left(-\frac{x}{2}\right)^2 \\ &= 729 + 6(243)\left(-\frac{x}{2}\right) + 15(81)\left(\frac{x^2}{4}\right) \\ &= 729 - 729x + \frac{1215}{4}x^2 \end{aligned}

The coefficient of one term

Find the coefficient of $x^3$ in the expansion of $(1 + 2x)^7$ .

Solution

The $x^3$ term has $r = 3$ :

\binom{7}{3}(1)^4(2x)^3 = 35 \times 8x^3 = 280x^3

The coefficient is $280$ .

The term independent of x

Find the term independent of $x$ in the expansion of $\left(x^2 + \dfrac{2}{x}\right)^6$ .

Solution

General term:

\binom{6}{r}\left(x^2\right)^{6-r}\left(\frac{2}{x}\right)^r = \binom{6}{r}2^r x^{12 - 2r}x^{-r} = \binom{6}{r}2^r x^{12 - 3r}

Independent of $x$ means the power is $0$ : $12 - 3r = 0$ , so $r = 4$ .

\binom{6}{4}2^4 = 15 \times 16 = 240

A coefficient in a product

Find the coefficient of $x^2$ in the expansion of $(1 + 3x)(2 - x)^5$ .

Solution

Expand $(2 - x)^5$ as far as $x^2$ :

(2 - x)^5 = 32 + 5(16)(-x) + 10(8)(-x)^2 + \cdots = 32 - 80x + 80x^2 + \cdots

The $x^2$ terms in the product come from $1 \times 80x^2$ and $3x \times (-80x)$ :

80 - 240 = -160

Two unknowns from two coefficients

In the expansion of $(1 + ax)^n$ , where $n$ is a positive integer, the coefficient of $x$ is $12$ and the coefficient of $x^2$ is $60$ . Find $a$ and $n$ .

Solution

(1 + ax)^n = 1 + n(ax) + \frac{n(n - 1)}{2}(ax)^2 + \cdots

So $na = 12$ and $\dfrac{n(n - 1)}{2}a^2 = 60$ .

Write the second as $\tfrac{1}{2}(na)(n - 1)a = 60$ and substitute $na = 12$ :

6(n - 1)a = 60 \quad\Rightarrow\quad (n - 1)a = 10 \quad\Rightarrow\quad na - a = 10 \quad\Rightarrow\quad 12 - a = 10

So $a = 2$ and $n = 6$ . Check: $(1 + 2x)^6 = 1 + 12x + 60x^2 + \cdots$ .

An approximation

(a) Find the first three terms in the expansion of $(2 - x)^6$ in ascending powers of $x$ .

(b) Use your answer to estimate $1.99^6$ , giving your answer to 3 decimal places.

Solution

(a)

(2 - x)^6 = 64 + 6(32)(-x) + 15(16)x^2 + \cdots = 64 - 192x + 240x^2 + \cdots

(b) $1.99 = 2 - 0.01$ , so put $x = 0.01$ :

64 - 192(0.01) + 240(0.0001) = 64 - 1.92 + 0.024 = 62.104

The next term, $-160x^3 = -0.00016$ , is too small to affect the third decimal place, so $1.99^6 \approx 62.104$ .

y = (2 - x)^6 y = 64 - 192x + 240x^2

The first three terms (the parabola) follow $y = (2 - x)^6$ closely when $x$ is small, which is why the approximation works for $x = 0.01$ . Further from $0$ the dropped terms matter and the curves separate.

Two terms needed from one expansion

(a) Find the term independent of $x$ and the coefficient of $x^2$ in the expansion of $\left(x - \dfrac{2}{x}\right)^6$ .

(b) Hence find the coefficient of $x^2$ in the expansion of $\left(1 + x^2\right)\left(x - \dfrac{2}{x}\right)^6$ .

Solution

(a) General term: $\binom{6}{r}x^{6-r}\left(-\dfrac{2}{x}\right)^r = \binom{6}{r}(-2)^r x^{6 - 2r}$ .

Independent of $x$ : $6 - 2r = 0$ , $r = 3$ : $\binom{6}{3}(-2)^3 = 20 \times (-8) = -160$ .

$x^2$ : $6 - 2r = 2$ , $r = 2$ : $\binom{6}{2}(-2)^2 = 15 \times 4 = 60$ .

(b) The $x^2$ term of the product comes from $1 \times (\text{the } x^2 \text{ term})$ and $x^2 \times (\text{the constant term})$ :

60 + (-160) = -100

Watch out

Missing brackets. $(2x)^3 = 8x^3$ , not $2x^3$ , and $\left(-\tfrac{x}{2}\right)^2 = \tfrac{x^2}{4}$ , not $-\tfrac{x^2}{2}$ . Always bracket $a$ and $b$ before raising to a power.

Losing the sign of $b$ . In $(2 - x)^5$ , $b = -x$ , so odd powers of $b$ are negative.

Off by one. The $x^3$ term uses $\binom{n}{3}$ and is the fourth term. "The third term" uses $\binom{n}{2}$ .

Forgetting the power of $a$ . In $(3 + 2x)^5$ , the $x^2$ term is $\binom{5}{2}3^3(2x)^2$ , not $\binom{5}{2}(2x)^2$ .

Coefficient versus term. "The coefficient of $x^3$ " is a number, such as $280$ . "The term in $x^3$ " is $280x^3$ .

Exam tip

Command words. "In ascending powers of $x$ " means start with the constant. "Up to and including the term in $x^3$ " means four terms. "Term independent of $x$ " means the constant term.
Show the structure. Write $\binom{6}{2}3^4\left(-\tfrac{x}{2}\right)^2$ before simplifying. If the arithmetic slips, the method mark is still earned.
Simplify fully. The accuracy mark needs each coefficient as an integer or a fraction in lowest terms.
Products. State which pairs of terms you are combining. A common 4-mark question: two marks for the binomial terms, one for combining, one for the answer.
Not required. The greatest term and properties of the coefficients are not on the syllabus. Expansions with negative or fractional $n$ belong to Paper 3.

Summary

$(a + b)^n = \sum \binom{n}{r}a^{n-r}b^r$ for a positive integer $n$ , with $n + 1$ terms.
$\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$ : the number of ways of choosing $r$ from $n$ ; the entries of Pascal's triangle.
$\binom{n}{r} = \binom{n}{n - r}$ , $\binom{n}{1} = n$ , $\binom{n}{2} = \tfrac{1}{2}n(n - 1)$ .
General term $\binom{n}{r}a^{n-r}b^r$ is the $(r + 1)$ th term; use it to find one term without expanding everything.
Bracket $a$ and $b$ , including signs and coefficients.
For a product, expand as far as needed and add the products of terms whose powers sum to the target.

Practice questions

Question

Expand $(1 - 2x)^5$ fully.
Find the first four terms of $(2 + 3x)^8$ in ascending powers of $x$ .
Evaluate $\binom{7}{3}$ , and show that $\binom{10}{4} = \binom{10}{6}$ .
Find the coefficient of $x^4$ in the expansion of $(1 + x)(1 - x)^6$ .
The coefficient of $x^3$ in the expansion of $(1 + kx)^6$ is $-160$ . Find $k$ .
Find the term independent of $x$ in the expansion of $\left(2x - \dfrac{1}{x^2}\right)^9$ .
(a) Find the first three terms of $(2 - x)^6$ in ascending powers of $x$ . (b) Hence find the coefficient of $x^2$ in $(3 + 2x)(2 - x)^6$ .
In the expansion of $(1 + ax)^5(1 - 2x)^4$ , the coefficient of $x$ is zero. Find $a$ , and hence find the coefficient of $x^2$ .
The term independent of $x$ in the expansion of $\left(x + \dfrac{k}{x}\right)^8$ , where $k$ is a positive constant, is $1120$ . Find $k$ , and the coefficient of $x^2$ in the expansion.

Answers

$a = 1$ , $b = -2x$ , coefficients $1, 5, 10, 10, 5, 1$ : $1 - 10x + 40x^2 - 80x^3 + 80x^4 - 32x^5$ .
$2^8 + 8(2^7)(3x) + 28(2^6)(3x)^2 + 56(2^5)(3x)^3 = 256 + 3072x + 16\,128x^2 + 48\,384x^3$ .
$\binom{7}{3} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ . $\binom{10}{4} = \dfrac{10!}{4!\,6!}$ and $\binom{10}{6} = \dfrac{10!}{6!\,4!}$ , which are the same expression; both equal $210$ .
In $(1 - x)^6$ , the $x^4$ coefficient is $\binom{6}{4} = 15$ and the $x^3$ coefficient is $-\binom{6}{3} = -20$ . In the product: $1 \times 15 + 1 \times (-20) = -5$ .
$\binom{6}{3}k^3 = 20k^3 = -160$ , so $k^3 = -8$ and $k = -2$ .
General term $\binom{9}{r}(2x)^{9-r}\left(-x^{-2}\right)^r = \binom{9}{r}2^{9-r}(-1)^r x^{9 - 3r}$ . $9 - 3r = 0$ gives $r = 3$ : $84 \times 2^6 \times (-1) = -5376$ .
(a) $64 - 192x + 240x^2$ . (b) $3 \times 240 + 2 \times (-192) = 720 - 384 = 336$ .
$(1 + ax)^5 = 1 + 5ax + 10a^2x^2 + \cdots$ and $(1 - 2x)^4 = 1 - 8x + 24x^2 + \cdots$ . Coefficient of $x$ : $5a - 8 = 0$ , so $a = \tfrac{8}{5}$ . Coefficient of $x^2$ : $24 + 5a(-8) + 10a^2 = 24 - 64 + 25.6 = -14.4 = -\tfrac{72}{5}$ .
General term $\binom{8}{r}x^{8-r}k^r x^{-r} = \binom{8}{r}k^r x^{8 - 2r}$ . Independent: $r = 4$ , $70k^4 = 1120$ , so $k^4 = 16$ and $k = 2$ . Coefficient of $x^2$ : $8 - 2r = 2$ , $r = 3$ : $\binom{8}{3}2^3 = 56 \times 8 = 448$ .

Where the coefficients come from

Why: counting choices

Factorials and (nr)\binom{n}{r}(rn​)

The binomial theorem

Products of two brackets

Worked examples

Practice questions

Console

Factorials and $\binom{n}{r}$