The binomial expansion · Sequences, series & binomial expansion

Multiplying out $(a + b)^2$ is quick, but $(2 - x)^5$ means five brackets. The binomial expansion writes down the answer directly. It uses the numbers ${}^nC_r$ from counting selections (see counting): the number of ways to choose $r$ of the $n$ brackets to give a $b$ .

Pascal’s triangle and ⁿCᵣ

The coefficients of $(a + b)^n$ are row $n$ of Pascal’s triangle. Each number is the sum of the two above it:

Pascal's triangleRow 5: 1, 5, 10, 10, 5, 1 are ⁵C₀ … ⁵C₅

The same numbers come from ${}^nC_r = \dfrac{n!}{r!\,(n - r)!}$ , which is quicker for large $n$ . For example, ${}^7C_3 = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ .

The general term

In $(a + b)^n$ , the powers of $a$ go down from $n$ to 0 while the powers of $b$ go up from 0 to $n$ . The term with $b^r$ is:

The (r + 1)th termⁿCᵣ aⁿ⁻ʳ bʳ: the powers always add up to n

When $b$ is something like $-x$ or $2x$ , put it in brackets and raise all of it to the power: ${(-x)^3 = -x^3}$ and ${(2x)^2 = 4x^2}$ . A minus sign in $b$ makes the signs alternate.

Build any expansion term by term:

Building a binomial expansionPick a bracket and n

(2 − x)⁵ = 32 − 80x + 80x² − 40x³ + 10x⁴ − x⁵

r	ⁿCᵣ	2ⁿ⁻ʳ	(−x)ʳ	term
0	1	32	1	32
1	5	16	−1	−80x
2	10	8	1	80x²
3	10	4	−1	−40x³
4	5	2	1	10x⁴
5	1	1	−1	−x⁵

1, 5, 10, 10, 5, 1row 5 of Pascal's triangle

n = 5

Each term is ⁿCᵣ × 2ⁿ⁻ʳ × (−x)ʳ. The powers of 2 go down as the powers of x go up, and the minus sign makes the signs alternate.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q3

Write down the binomial expansion of $(2 - x)^5$ in ascending powers of $x$ .

Use your expansion in (a) to evaluate $(1.98)^5$ correct to four decimal places.

The coefficients
- Row 5: ${1, 5, 10, 10, 5, 1}$ .
- Here ${a = 2}$ and ${b = -x}$ .
Think first. Row 5 of Pascal's triangle?
Each term
- ${r = 0}$ : ${1 \times 32 = 32}$ .
- ${r = 1}$ : ${5 \times 16 \times (-x) = -80x}$ .
- ${r = 2}$ : ${10 \times 8 \times x^2 = 80x^2}$ .
- ${r = 3}$ : ${10 \times 4 \times (-x^3) = -40x^3}$ .
- ${r = 4}$ : ${5 \times 2 \times x^4 = 10x^4}$ .
- ${r = 5}$ : ${1 \times 1 \times (-x^5) = -x^5}$ .
- Put the terms together:
$\begin{aligned} (2 - x)^5 = {} & 32 - 80x + 80x^2 \\ & - 40x^3 + 10x^4 - x^5 \end{aligned}$
Think first. Work out each term ⁵Cᵣ 2⁵⁻ʳ(−x)ʳ.
Choose x for (1.98)⁵
- ${2 - x = 1.98}$ , so ${x = 0.02}$ .
Think first. 2 − x = 1.98, so x = ?
Substitute
- ${32 - 80(0.02) = 32 - 1.6 = 30.4}$ .
- ${+\ 80(0.02)^2 = +\ 0.032}$ , giving ${30.432}$ .
- ${-\ 40(0.02)^3 = -\ 0.00032}$ , giving ${30.43168}$ .
- ${+\ 10(0.02)^4 = +\ 0.0000016}$ , giving ${30.4316816}$ .
- The last term is far too small to matter: ${(1.98)^5 \approx 30.4317}$ .
Think first. Work out the first few terms with x = 0.02.

Approximating a power

To estimate something like $(1.02)^5$ , match it to an expansion you know. Choose $x$ so that the bracket equals the number, then add the first few terms. Because $x$ is small, each term is much smaller than the one before, so a few terms give plenty of accuracy.

One coefficient, and unknowns

You don’t need the whole expansion to find one term. Use the general term with the right $r$ . And when a question gives some coefficients, write each as an expression and set up equations.

Worked example · WAEC 2012

WAEC 2012 · Paper 2 · Q9

The first three terms of the expansion of $(1 + mx)^n$ in ascending powers of $x$ are $1 + 14x + 84x^2$ . Find the values of $m$ and $n$ .

Using the values of $m$ and $n$ obtained in (a), calculate, correct to three significant figures, the value of $(1.06)^n$ .

Write the first three terms
- The first two terms: ${1 + n(mx)}$ .
- The third term: ${\frac{n(n - 1)}{2}(mx)^2 = \frac{n(n - 1)}{2}m^2x^2}$ .
Think first. The general term with b = mx.
Match the coefficients
- The $x$ terms: ${nm = 14}$ , so ${m = \frac{14}{n}}$ .
- The $x^2$ terms: ${\frac{n(n - 1)}{2}m^2 = 84}$ .
Think first. Compare with 1 + 14x + 84x².
Solve
- ${\frac{n(n - 1)}{2} \times \frac{196}{n^2} = 84}$ .
- Cancel one $n$ : ${\frac{98(n - 1)}{n} = 84}$ .
- Multiply by $n$ : ${98n - 98 = 84n}$ .
- So ${14n = 98}$ , and ${n = 7}$ . Then ${m = \frac{14}{7} = 2}$ .
Think first. Substitute m = 14/n into the second equation.
Approximate (1.06)⁷
- ${2x = 0.06}$ , so ${x = 0.03}$ .
- ${(1 + 2x)^7 = 1 + 14x + 84x^2 + 280x^3 + \ldots}$
- The terms are ${1}$ , ${0.42}$ , ${0.0756}$ , ${0.00756}$ , and then tiny ones.
- Add them: ${1.503\ldots}$
- So ${(1.06)^7 \approx 1.50}$ to three significant figures.
Think first. 1 + 2x = 1.06, so x = ?

More: coefficients and unknowns

More: expansions and approximations

Fractional and negative powers

The same expansion works when $n$ is a fraction such as $\frac13$ or a negative number such as $-1$ , but only for $(1 + x)^n$ , with 1 first in the bracket. There is no row of Pascal’s triangle for $n = \frac12$ , so write each coefficient from $n$ itself:

\begin{aligned} (1 + x)^n = 1 &+ nx + \frac{n(n - 1)}{2!}x^2 \\ &+ \frac{n(n - 1)(n - 2)}{3!}x^3 + \ldots \end{aligned}

Now the brackets $n - 1$ , $n - 2$ , … never reach 0, so the series never stops. It only adds up to the right value when $x$ is small:

\text{valid for } |x| < 1

Close only for small x√(1 + x) and its first three terms agree inside |x| < 1 and pull apart outside it

When the bracket is $(1 + bx)$ , put $bx$ in place of $x$ everywhere, and the condition becomes $|bx| < 1$ .

Worked example

Expand

(1 - 4x)^{\frac12}

in ascending powers of

x

up to the term in

x^2

, and state the values of

x

for which it is valid. Use it to find

\sqrt{0.96}

, correct to four decimal places.

Name n and the x
- ${n = \frac12}$ .
- Put $-4x$ in place of $x$ , in brackets.
Think first. What are n and the term that replaces x?
The x term
- ${\frac12 \times (-4x) = -2x}$ .
Think first. Work out n × (−4x).
The x² term
- The coefficient: ${\frac{\frac12 \times \left(-\frac12\right)}{2} = -\frac18}$ .
- Square the whole bracket: ${(-4x)^2 = 16x^2}$ .
- Multiply: ${-\frac18 \times 16x^2 = -2x^2}$ .
- So ${(1 - 4x)^{\frac12} \approx 1 - 2x - 2x^2}$ .
Think first. Work out n(n − 1) ÷ 2! first, then (−4x)².
When is it valid?
- ${|-4x| < 1}$ .
- Divide by 4: ${|x| < \frac14}$ .
Think first. The condition is on the whole of −4x.
Choose x for √0.96
- ${1 - 4x = 0.96}$ , so ${4x = 0.04}$ .
- So ${x = 0.01}$ , which is inside ${|x| < \frac14}$ .
Think first. 1 − 4x = 0.96, so x = ?
Substitute
- ${1 - 2(0.01) - 2(0.01)^2}$ .
- ${= 1 - 0.02 - 0.0002}$ .
- So ${\sqrt{0.96} \approx 0.9798}$ .

More: fractional and negative powers

WAEC 2009 · Paper 2 · Q4Write down the first three terms of the binomial expansion of $(1 + 3x)^{\frac13}$ in ascending powers of $x$ .

Your turn

WAEC 2022 · Paper 2 · Q11

Sequences, series & binomial expansion

(a)
Find the binomial expansion of $(1 + 2x)^7$ and $(1 - 2x)^7$ .
Show the answer
$(1 \pm 2x)^7 = 1 \pm 14x + 84x^2 \pm 280x^3 + 560x^4 \pm 672x^5 + 448x^6 \pm 128x^7$
(b)
Using the result in (a), find, correct to three decimal places, the value of $(1.2)^7 - (0.8)^7$ .

Worked solution (try it first)

(a)

Each term of $(1 + 2x)^7$ is $\binom7r(2x)^r$ : $1 + 14x + 84x^2 + 280x^3 + 560x^4 + 672x^5 + 448x^6 + 128x^7$ .
$(1 - 2x)^7$ has the same terms with the odd powers of $x$ negative: $1 - 14x + 84x^2 - 280x^3 + 560x^4 - 672x^5 + 448x^6 - 128x^7$ .

(b)

$(1.2)^7 - (0.8)^7$ is $(1 + 2x)^7 - (1 - 2x)^7$ with $x = 0.1$ .
Subtracting, the even powers cancel and the odd ones double: $28x + 560x^3 + 1344x^5 + 256x^7$ .
With $x = 0.1$ : $2.8 + 0.56 + 0.01344 + 0.0000256 = 3.3734656$ , so the value is $3.373$ to three decimal places.

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