Year 1 Pure · Chapter 7

Binomial Expansion

Specification: Edexcel Pure 4 — Sequences and series · AQA section D

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In this chapter

1. Pascal’s triangle
2. The binomial coefficient
3. The binomial theorem
4. Estimating with the binomial expansion
Practice

Key facts — this chapter

$(a + b)^n$ expands into $n + 1$ terms; the powers of $a$ fall from $n$ to $0$ while the powers of $b$ rise from $0$ to $n$ .
The coefficients are row $n$ of Pascal’s triangle, where each entry is the sum of the two above it.
Formally, the coefficient of the term in $b^r$ is $\dbinom{n}{r} = \dfrac{n!}{r!\,(n-r)!}$ , often written $^nC_r$ .
Binomial theorem: $(a+b)^n = \displaystyle\sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$ .
For $(1 + x)^n$ the terms are simply $1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots$
Putting a small number in for $x$ gives a fast, accurate estimate of a power — the more terms, the closer.

Multiplying out $(a+b)^2$ is easy; $(a+b)^7$ by hand is misery. The binomial expansion is the shortcut — a pattern that hands you every coefficient without multiplying a single bracket.

1. Pascal’s triangle

Expand $(a+b)^n$ for $n = 0, 1, 2, 3$ and the coefficients line up into a famous triangle.

Key result

Each row of Pascal’s triangle gives the coefficients of $(a+b)^n$ . Every entry is the sum of the two directly above it, with $1$ s down each edge:

\begin{array}{c} 1 \\ 1 \quad 1 \\ 1 \quad 2 \quad 1 \\ 1 \quad 3 \quad 3 \quad 1 \\ 1 \quad 4 \quad 6 \quad 4 \quad 1 \end{array}

Pascal's triangle rows 0 to 5, each entry the sum of the two above, with row 4 (1 4 6 4 1) highlighted — Each entry is the sum of the two above it; row 4 gives the coefficients of (a + b)⁴.

Alongside the coefficients, the powers follow their own tidy rule: $a$ counts down, $b$ counts up, and every term’s powers add to $n$ .

The five terms of (a+b)^4 with the powers of a falling 4 to 0 and powers of b rising 0 to 4 — In (a + b)⁴: a's power falls 4→0, b's rises 0→4 — each term totalling 4.

Worked example 7.1 · Expanding with Pascal's triangle

Expand $(a + b)^4$ .

Row 4 of the triangle is $1, 4, 6, 4, 1$ .
Pair each coefficient with falling powers of $a$ and rising powers of $b$ :
$(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4$

Worked example 7.2 · A binomial with a coefficient inside

Expand $(2 + x)^3$ .

Row 3 is $1, 3, 3, 1$ . Here $a = 2$ and $b = x$ :
$(2+x)^3 = 1\cdot 2^3 + 3\cdot 2^2 x + 3 \cdot 2 \, x^2 + x^3$
Work out each coefficient:
$(2+x)^3 = 8 + 12x + 6x^2 + x^3$

Common mistake

When a term like $2$ or $3x$ sits inside the bracket, the power hits all of it: the second term of $(2+x)^3$ is $3 \cdot 2^2 \cdot x = 12x$ , not $3x$ . Raise the whole thing, coefficient and all.

2. The binomial coefficient

Pascal’s triangle is perfect for small $n$ , but writing out twelve rows to expand $(a+b)^{12}$ is no fun. The formula $^nC_r$ jumps straight to any coefficient.

Key result

The coefficient of the term containing $b^r$ in $(a+b)^n$ is

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

where $n! = n \times (n-1) \times \dots \times 2 \times 1$ . It counts the ways of choosing which $r$ of the $n$ brackets contribute a $b$ .

Why it works

Expanding $(a+b)^n$ means multiplying $n$ brackets and collecting like terms. A term with $b^r$ comes from choosing $b$ out of exactly $r$ of the brackets (and $a$ from the rest) — and the number of ways to make that choice is $\binom{n}{r}$ . That count is the coefficient.

Step n up and down. Watch Pascal's row grow (each entry the sum of the two above), and see it become the coefficients of (a + b)ⁿ below.

Worked example 7.3 · A single term via the formula

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

For $(1+x)^7$ the term in $x^3$ has coefficient $\binom{7}{3}$ :
$\binom{7}{3} = \frac{7!}{3!\,4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$

So the coefficient of $x^3$ is $35$ .

3. The binomial theorem

Putting it together gives the general formula — and for $(1+x)^n$ it’s especially clean.

Key result

(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r

In the common case $a = 1$ , $b = x$ :

(1+x)^n = 1 + nx + \binom{n}{2}x^2 + \binom{n}{3}x^3 + \dots + x^n

Worked example 7.4 · First terms of an expansion

Find the first four terms, in ascending powers of $x$ , of $(1 + 2x)^6$ .

Use $(1+x)^n$ with the " $x$ " being $2x$ and $n = 6$ . The coefficients are $\binom{6}{0}, \binom{6}{1}, \binom{6}{2}, \binom{6}{3} = 1, 6, 15, 20$ .
Attach the powers of $2x$ :
$1 + 6(2x) + 15(2x)^2 + 20(2x)^3 + \dots$
Simplify each — remember the power hits the $2$ as well:
$1 + 12x + 60x^2 + 160x^3 + \dots$

Common mistake

$15(2x)^2$ is $15 \cdot 4x^2 = 60x^2$ , not $30x^2$ — square the whole of $2x$ . Forgetting to raise the inner coefficient is the classic binomial error, and it quietly wrecks every term after the first.

4. Estimating with the binomial expansion

A neat payoff: choosing a small value of $x$ turns an expansion into a quick, surprisingly accurate numerical estimate.

The idea

To estimate a power like $(1.02)^8$ , write the base as $1 + x$ with $x$ small, expand $(1+x)^n$ , and keep the first few terms. Because each term carries a higher power of a small number, the later terms are tiny — so a handful of terms is plenty.

Worked example 7.5 · Estimating a power

Use the first three terms of the expansion of $(1 + x)^8$ to estimate $(1.02)^8$ .

The first three terms of $(1+x)^8$ are
$1 + 8x + \binom{8}{2}x^2 = 1 + 8x + 28x^2$
Here $1.02 = 1 + 0.02$ , so substitute $x = 0.02$ :
$1 + 8(0.02) + 28(0.02)^2 = 1 + 0.16 + 0.0112 = 1.1712$

The true value is $1.171659\ldots$ , so three terms already give four-figure accuracy — the next term, in $x^3$ , contributes only about $0.0004$ .

Practice

Twelve questions in three tiers. Try each one properly before opening the solution — the solutions show full method, not just answers.

Warm-up

1 [2 marks]

Write down row 5 of Pascal’s triangle, and hence the coefficients of $(a + b)^5$ .

Show solution

Each entry is the sum of the two above it. Row 5 is

1, \;5, \;10, \;10, \;5, \;1

These are exactly the coefficients of $(a+b)^5$ .

2 [3 marks]

Expand $(a + b)^3$ and use it to expand $(x + 2)^3$ .

Show solution

Row 3 is $1, 3, 3, 1$ , so $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ .

With $a = x$ , $b = 2$ :

(x+2)^3 = x^3 + 3x^2(2) + 3x(2)^2 + 2^3 = x^3 + 6x^2 + 12x + 8

3 [3 marks]

Evaluate:

$\dbinom{6}{2}$
$\dbinom{6}{0}$
$\dbinom{8}{8}$

Show solution

(a) $\dbinom{6}{2} = \dfrac{6 \times 5}{2 \times 1} = 15$ .

(b) $\dbinom{6}{0} = 1$ (one way to choose nothing).

(c) $\dbinom{8}{8} = 1$ (one way to choose everything).

4 [2 marks]

Write down the number of terms in the expansion of $(a + b)^{10}$ , and the powers of $a$ and $b$ in its third term.

Show solution

$(a+b)^{10}$ has $10 + 1 = 11$ terms.

Counting from the first term $a^{10}$ , the third term is the one in $b^2$ : it contains $a^{8}b^{2}$ (the powers always sum to $10$ ).

Standard

5 [3 marks]

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

Show solution

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

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

6 [3 marks]

Find the coefficient of $x^4$ in the expansion of $(1 + x)^{10}$ .

Show solution

The coefficient of $x^4$ is $\dbinom{10}{4}$ :

\binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = \frac{5040}{24} = 210

7 [4 marks]

Find the first three terms, in ascending powers of $x$ , of $(1 + 3x)^5$ .

Show solution

Coefficients $\binom{5}{0}, \binom{5}{1}, \binom{5}{2} = 1, 5, 10$ , attached to powers of $3x$ :

1 + 5(3x) + 10(3x)^2 + \dots = 1 + 15x + 90x^2 + \dots

(Note $10(3x)^2 = 10 \times 9x^2 = 90x^2$ .)

8 [4 marks]

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

Show solution

The term in $x^2$ takes $b = 2x$ from $2$ of the brackets and $a = 3$ from the other $3$ :

\binom{5}{2}\,3^{3}\,(2x)^2 = 10 \times 27 \times 4x^2 = 1080x^2

So the coefficient of $x^2$ is $1080$ .

9 [3 marks]

Use the first three terms of the expansion of $(1 + x)^{10}$ to estimate $(1.01)^{10}$ .

Show solution

First three terms: $1 + 10x + \binom{10}{2}x^2 = 1 + 10x + 45x^2$ .

Substitute $x = 0.01$ :

1 + 10(0.01) + 45(0.01)^2 = 1 + 0.1 + 0.0045 = 1.1045

(The true value is $1.10462\ldots$ )

Exam style

10 [5 marks]

Find the first four terms, in ascending powers of $x$ , of $(1 - 2x)^7$ . [3]
Hence estimate $(0.98)^7$ . [2]

Show solution

(a) Coefficients $\binom{7}{0}, \binom{7}{1}, \binom{7}{2}, \binom{7}{3} = 1, 7, 21, 35$ , attached to powers of $(-2x)$ :

1 + 7(-2x) + 21(-2x)^2 + 35(-2x)^3 + \dots

= 1 - 14x + 84x^2 - 280x^3 + \dots

(b) $0.98 = 1 - 2(0.01)$ , so substitute $x = 0.01$ :

\begin{gathered} 1 - 14(0.01) + 84(0.01)^2 - 280(0.01)^3 \\[0.3em] = 1 - 0.14 + 0.0084 - 0.00028 = 0.86812 \end{gathered}

(The true value is $0.868125\ldots$ )

11 [5 marks]

In the expansion of $(1 + kx)^6$ , the coefficient of $x^2$ is $60$ . Given that $k > 0$ , find the value of $k$ .

Show solution

The coefficient of $x^2$ is $\binom{6}{2}k^2$ :

\binom{6}{2}k^2 = 15k^2

Set equal to $60$ :

15k^2 = 60 \;\;\Longrightarrow\;\; k^2 = 4 \;\;\Longrightarrow\;\; k = 2 \quad (k > 0)

12 [5 marks]

The first three terms of $(1 + ax)^n$ , in ascending powers of $x$ , are $1 + 12x + 60x^2$ . Find the values of $a$ and $n$ .

Show solution

The $x$ -term gives $\binom{n}{1}a = na = 12$ .
The $x^2$ -term gives $\binom{n}{2}a^2 = \dfrac{n(n-1)}{2}a^2 = 60$ .
From (1), $a = \dfrac{12}{n}$ . Substitute into (2):
$\frac{n(n-1)}{2} \cdot \frac{144}{n^2} = 60 \;\;\Longrightarrow\;\; \frac{72(n-1)}{n} = 60$
Solve: $72(n-1) = 60n \Rightarrow 72n - 72 = 60n \Rightarrow 12n = 72 \Rightarrow n = 6$ .
Then $a = \dfrac{12}{6} = 2$ .

So $a = 2$ and $n = 6$ .