Year 2 Pure · Chapter 7

The Binomial Series

Specification: Edexcel Pure 4.1 — Sequences and series · AQA section D1

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In this chapter

1. The binomial series for any power
2. Expanding (1 + kx) to a power
3. Expanding (a + bx) to a power
4. Approximations
5. Using partial fractions
Practice

Key facts — this chapter

For any rational $n$ and $|x| < 1$ : $(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \dots$
When $n$ is a positive whole number the series stops. Otherwise it goes on for ever, and is only valid for $|x| < 1$ .
For $(1 + kx)^n$ , replace $x$ by $kx$ throughout. It is valid for $|kx| < 1$ , that is $|x| < \frac{1}{|k|}$ .
For $(a + bx)^n$ , take out $a^n$ first: $a^n\left(1 + \frac{b}{a}x\right)^n$ , valid for $|x| < \left|\frac{a}{b}\right|$ .
To approximate a value, choose $x$ small and inside the interval of validity, then use the first few terms.
For a fraction, split it into partial fractions, expand each one, and add. The result is valid where both expansions are.

In Year 1 you expanded $(1 + x)^n$ for positive whole numbers $n$ , and the expansion stopped after $n + 1$ terms. The same formula works for negative and fractional powers too, but then it never stops. That turns $\frac{1}{1 + x}$ and $\sqrt{1 + x}$ into series you can use for approximations, as long as $x$ is small enough.

1. The binomial series for any power

Key result

For any rational number $n$ , and $|x| < 1$ ,

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

If $n$ is a positive whole number, a factor $(n - n)$ eventually appears, every later term is $0$ , and the expansion is the finite one from Year 1. For any other $n$ it is an infinite series.

Worked example 7.1 · A negative power

Expand $(1 + x)^{-1}$ in ascending powers of $x$ up to the term in $x^3$ .

With $n = -1$ , the coefficients are

\begin{aligned} nx &= -x \\ \tfrac{(-1)(-2)}{2}x^2 &= x^2 \\ \tfrac{(-1)(-2)(-3)}{6}x^3 &= -x^3 \end{aligned}

So $(1 + x)^{-1} = 1 - x + x^2 - x^3 + \dots$ , valid for $|x| < 1$ . It’s the geometric series with first term $1$ and ratio $-x$ , which converges exactly when $|x| < 1$ .

Worked example 7.2 · A square root

Expand $\sqrt{1 + x}$ up to the term in $x^3$ .

Write it as $(1 + x)^{\frac{1}{2}}$ , so $n = \tfrac{1}{2}$ .
Work out each coefficient:
$\begin{aligned} \tfrac{\frac{1}{2}\left(-\frac{1}{2}\right)}{2} &= -\tfrac{1}{8} \\ \tfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6} &= \tfrac{1}{16} \end{aligned}$
So
$\begin{aligned} \sqrt{1 + x} &= 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 \\ &\quad + \tfrac{1}{16}x^3 - \dots \end{aligned}$
valid for $|x| < 1$ .

Why the terms must shrink

Each term carries a power of $x$ . When $|x| < 1$ the powers shrink, and the terms become small enough for the series to settle on a value. When $|x| > 1$ the powers grow and the series can’t settle. Try $x = 2$ in Example 7.1: the series $1 - 2 + 4 - 8 + \dots$ swings further and further, although $(1 + 2)^{-1} = \tfrac{1}{3}$ .

√(1 + x) with its 3-, 6- and 12-term approximations. Inside |x| < 1, more terms fit better; outside, more terms make it worse.

The solid curve is (1 + x)ⁿ; the dashed one is the series cut off after a number of terms. Add terms, then move x outside the shaded band |x| < 1 and watch what extra terms do there.

Power Terms

2. Expanding (1 + kx) to a power

Replace every $x$ in the formula by $kx$ , brackets and all. The series is valid while $|kx| < 1$ .

Worked example 7.3 · A cube in the denominator

Expand $\dfrac{1}{(1 - 2x)^3}$ up to the term in $x^3$ , and state the values of $x$ for which it is valid.

Write it as $(1 - 2x)^{-3}$ : $n = -3$ , with $-2x$ in place of $x$ .
Term by term, keeping the brackets:
$\begin{aligned} &1 + (-3)(-2x) + \tfrac{(-3)(-4)}{2}(-2x)^2 \\ &\quad + \tfrac{(-3)(-4)(-5)}{6}(-2x)^3 \\ &= 1 + 6x + 24x^2 + 80x^3 + \dots \end{aligned}$
It is valid for $|2x| < 1$ , that is $|x| < \tfrac{1}{2}$ .

Common mistake

Put the whole of $kx$ in a bracket before you raise it to a power: $(-2x)^2 = 4x^2$ and $(-2x)^3 = -8x^3$ . Writing $-2x^2$ gets both the number and the sign wrong.

Worked example 7.4 · Finding the constants

In the expansion of $(1 + ax)^n$ , the coefficient of $x$ is $-6$ and the coefficient of $x^2$ is $27$ . Find $a$ and $n$ , and state the values of $x$ for which the expansion is valid.

Write down the two coefficients:
$\begin{aligned} na &= -6 \\ \tfrac{n(n - 1)}{2}a^2 &= 27 \end{aligned}$
Square the first, $n^2a^2 = 36$ , and double the second, $n(n - 1)a^2 = 54$ . Divide the second by the first:
$\frac{n - 1}{n} = \frac{3}{2} \;\Rightarrow\; n = -2$
Then $a = \dfrac{-6}{-2} = 3$ . The expansion is of $(1 + 3x)^{-2}$ , valid for $|3x| < 1$ , that is $|x| < \tfrac{1}{3}$ .

3. Expanding (a + bx) to a power

The formula needs a $1$ at the front of the bracket. So take $a$ out first, remembering that it comes out raised to the power $n$ .

Key result

(a + bx)^n = a^n\left(1 + \frac{b}{a}x\right)^n

Expand the bracket as before, with $\frac{b}{a}x$ in place of $x$ , then multiply every term by $a^n$ . The result is valid for $\left|\frac{b}{a}x\right| < 1$ .

Worked example 7.5 · Taking out the 4

Expand $\sqrt{4 + x}$ up to the term in $x^3$ , and state the values of $x$ for which it is valid.

Take out $4^{\frac{1}{2}} = 2$ :
$\sqrt{4 + x} = 2\left(1 + \frac{x}{4}\right)^{\frac{1}{2}}$
Use Example 7.2 with $\frac{x}{4}$ in place of $x$ :
$\begin{aligned} &2\left(1 + \frac{x}{8} - \frac{x^2}{128} + \frac{x^3}{1024}\right) \\ &= 2 + \frac{x}{4} - \frac{x^2}{64} + \frac{x^3}{512} - \dots \end{aligned}$
It is valid for $\left|\frac{x}{4}\right| < 1$ , that is $|x| < 4$ .

Common mistake

The $4$ you take out is raised to the power too. So $\sqrt{4 + x} = 2\left(1 + \frac{x}{4}\right)^{\frac{1}{2}}$ , not $4\left(1 + \frac{x}{4}\right)^{\frac{1}{2}}$ , which is twice as big.

4. Approximations

The first few terms of a series make a good approximation when $x$ is small, because the later terms carry high powers of a small number.

Worked example 7.6 · Estimating a square root

Use the expansion in Example 7.5 to estimate $\sqrt{4.1}$ .

$\sqrt{4.1} = \sqrt{4 + x}$ with $x = 0.1$ , which is well inside $|x| < 4$ .
Substitute into the four terms:
$\begin{aligned} &2 + 0.025 - 0.00015625 \\ &\quad + 0.000001953125 \\ &= 2.024845703\ldots \end{aligned}$
The true value is $\sqrt{4.1} = 2.024845673\ldots$ , so four terms are already correct to $7$ decimal places.

When the number isn’t of the form $a + x$ with $x$ small, look for a value of $x$ that turns the bracket into something related to it.

Worked example 7.7 · Approximating √2

Expand $\sqrt{1 - 2x}$ up to the term in $x^3$ . By substituting $x = 0.01$ , find an approximation to $\sqrt{2}$ .

Use Example 7.2 with $-2x$ in place of $x$ :
$\begin{aligned} &1 + \tfrac{1}{2}(-2x) - \tfrac{1}{8}(-2x)^2 \\ &\quad + \tfrac{1}{16}(-2x)^3 \\ &= 1 - x - \tfrac{1}{2}x^2 - \tfrac{1}{2}x^3 - \dots \end{aligned}$
valid for $|x| < \tfrac{1}{2}$ , so $x = 0.01$ is allowed.
When $x = 0.01$ , the left side is $\sqrt{0.98} = \sqrt{\tfrac{49 \times 2}{100}} = \tfrac{7}{10}\sqrt{2}$ , and the series gives $1 - 0.01 - 0.00005 - 0.0000005 = 0.9899495$ .
So
$\begin{aligned} \sqrt{2} &\approx \frac{10}{7} \times 0.9899495 \\ &= 1.4142136 \text{ (7 d.p.)} \end{aligned}$
which agrees with $\sqrt{2} = 1.41421356\ldots$ to $7$ decimal places.

5. Using partial fractions

A fraction with two factors underneath is easiest to expand in pieces: split it into partial fractions, expand each, and add.

Worked example 7.8 · Expanding a fraction

Given that $\dfrac{1 + 5x}{(1 - x)(1 + 2x)} \equiv \dfrac{2}{1 - x} - \dfrac{1}{1 + 2x}$ , expand it up to the term in $x^3$ and state the values of $x$ for which it is valid.

Expand each partial fraction up to the term in $x^3$ :
$\begin{aligned} \frac{2}{1 - x} &= 2 + 2x + 2x^2 + 2x^3 \\ \frac{1}{1 + 2x} &= 1 - 2x + 4x^2 - 8x^3 \end{aligned}$
Subtract:
$\begin{aligned} &\frac{1 + 5x}{(1 - x)(1 + 2x)} \\ &= 1 + 4x - 2x^2 + 10x^3 + \dots \end{aligned}$
The first series is valid for $|x| < 1$ and the second for $|x| < \tfrac{1}{2}$ . Both hold when $|x| < \tfrac{1}{2}$ , so that is where the expansion is valid.

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 [3 marks]

Expand $(1 + x)^{-2}$ in ascending powers of $x$ up to the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Show solution

With $n = -2$ :

\begin{aligned} &1 + (-2)x + \tfrac{(-2)(-3)}{2}x^2 \\ &\quad + \tfrac{(-2)(-3)(-4)}{6}x^3 \\ &= 1 - 2x + 3x^2 - 4x^3 + \dots \end{aligned}

It is valid for $|x| < 1$ .

2 [4 marks]

Expand $\sqrt[3]{1 + 3x}$ up to the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Show solution

Write it as $(1 + 3x)^{\frac{1}{3}}$ : $n = \tfrac{1}{3}$ , with $3x$ in place of $x$ .

\begin{aligned} &1 + \tfrac{1}{3}(3x) + \tfrac{\frac{1}{3}\left(-\frac{2}{3}\right)}{2}(3x)^2 \\ &\quad + \tfrac{\frac{1}{3}\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{6}(3x)^3 \\ &= 1 + x - x^2 + \tfrac{5}{3}x^3 - \dots \end{aligned}

It is valid for $|3x| < 1$ , that is $|x| < \tfrac{1}{3}$ .

3 [2 marks]

State the values of $x$ for which the binomial expansion of each of these is valid.

$(1 - 4x)^{\frac{1}{2}}$ [1]
$(3 + 2x)^{-2}$ [1]

Show solution

(a) $|4x| < 1$ , that is $|x| < \tfrac{1}{4}$ .

(b) $(3 + 2x)^{-2} = 3^{-2}\left(1 + \tfrac{2}{3}x\right)^{-2}$ , so it is valid for $\left|\tfrac{2}{3}x\right| < 1$ , that is $|x| < \tfrac{3}{2}$ .

4 [3 marks]

Expand $\dfrac{1}{2 + x}$ up to the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Show solution

Take out $2^{-1} = \tfrac{1}{2}$ : $\dfrac{1}{2 + x} = \tfrac{1}{2}\left(1 + \tfrac{x}{2}\right)^{-1}$ .

\begin{aligned} &\tfrac{1}{2}\left(1 - \tfrac{x}{2} + \tfrac{x^2}{4} - \tfrac{x^3}{8}\right) \\ &= \tfrac{1}{2} - \tfrac{1}{4}x + \tfrac{1}{8}x^2 - \tfrac{1}{16}x^3 + \dots \end{aligned}

It is valid for $\left|\tfrac{x}{2}\right| < 1$ , that is $|x| < 2$ .

Standard

5 [6 marks]

Expand $(1 - x)^{-\frac{1}{2}}$ up to the term in $x^3$ . [3]
By substituting $x = 0.1$ , find an approximation to $\dfrac{1}{\sqrt{0.9}}$ , giving your answer to 5 decimal places. [1]
Hence find an approximation to $\sqrt{10}$ . [2]

Show solution

(a) $n = -\tfrac{1}{2}$ , with $-x$ in place of $x$ :

\begin{aligned} &1 + \left(-\tfrac{1}{2}\right)(-x) + \tfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2}(-x)^2 \\ &\quad + \tfrac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{6}(-x)^3 \\ &= 1 + \tfrac{1}{2}x + \tfrac{3}{8}x^2 + \tfrac{5}{16}x^3 + \dots \end{aligned}

(b) $1 + 0.05 + 0.00375 + 0.0003125 = 1.0540625$ , which is $1.05406$ to 5 decimal places.

(c) $\dfrac{1}{\sqrt{0.9}} = \sqrt{\dfrac{10}{9}} = \dfrac{\sqrt{10}}{3}$ , so

\begin{aligned} \sqrt{10} &\approx 3 \times 1.0540625 \\ &= 3.16219 \text{ (5 d.p.)} \end{aligned}

The true value is $3.16228$ to 5 decimal places.

6 [4 marks]

Expand $\dfrac{1 + 2x}{(1 - x)^2}$ in ascending powers of $x$ up to the term in $x^3$ , and state the values of $x$ for which the expansion is valid.

Show solution

$(1 - x)^{-2} = 1 + 2x + 3x^2 + 4x^3 + \dots$ (Question 1 with $-x$ in place of $x$ ). Multiply by $1 + 2x$ , keeping terms up to $x^3$ :

\begin{aligned} &1 + 2x + 3x^2 + 4x^3 \\ &\quad + 2x + 4x^2 + 6x^3 + \dots \\ &= 1 + 4x + 7x^2 + 10x^3 + \dots \end{aligned}

It is valid for $|x| < 1$ .

7 [6 marks]

Given that $(1 + kx)^n = 1 + 2x - 2x^2 + \dots$ , where $n$ is not a positive integer,

find the values of $k$ and $n$ [4]
find the coefficient of $x^3$ [1]
state the values of $x$ for which the expansion is valid. [1]

Show solution

(a) $nk = 2$ and $\tfrac{n(n - 1)}{2}k^2 = -2$ . So $n^2k^2 = 4$ and $n(n - 1)k^2 = -4$ . Dividing,

\frac{n - 1}{n} = -1 \;\Rightarrow\; n = \tfrac{1}{2}

and $k = \dfrac{2}{n} = 4$ .

(b)

\tfrac{\frac{1}{2}\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{6} \times 4^3 = \tfrac{1}{16} \times 64 = 4

(c) $|4x| < 1$ , that is $|x| < \tfrac{1}{4}$ .

8 [8 marks]

f(x) = \frac{4 - 5x}{(1 + x)(2 - x)}

Express $f(x)$ in partial fractions. [3]
Hence find the expansion of $f(x)$ in ascending powers of $x$ up to the term in $x^2$ . [4]
State the values of $x$ for which the expansion is valid. [1]

Show solution

(a) $4 - 5x \equiv A(2 - x) + B(1 + x)$ . Put $x = -1$ : $9 = 3A$ , so $A = 3$ . Put $x = 2$ : $-6 = 3B$ , so $B = -2$ .

f(x) = \frac{3}{1 + x} - \frac{2}{2 - x}

(b) Expand each part. For the second, take out the $2$ : $\dfrac{2}{2 - x} = \left(1 - \dfrac{x}{2}\right)^{-1}$ .

\begin{aligned} \frac{3}{1 + x} &= 3 - 3x + 3x^2 - \dots \\ \frac{2}{2 - x} &= 1 + \tfrac{1}{2}x + \tfrac{1}{4}x^2 + \dots \end{aligned}

Subtracting, $f(x) = 2 - \tfrac{7}{2}x + \tfrac{11}{4}x^2 + \dots$

(c) The first expansion needs $|x| < 1$ and the second $|x| < 2$ . Both hold when $|x| < 1$ .

9 [6 marks]

Expand $\sqrt[3]{8 + x}$ up to the term in $x^2$ . [3]
Use your expansion to estimate $\sqrt[3]{9}$ , to 4 decimal places. [2]
Explain why the expansion should not be used with $x = 19$ to estimate $\sqrt[3]{27}$ . [1]

Show solution

(a) Take out $8^{\frac{1}{3}} = 2$ :

\begin{aligned} &2\left(1 + \tfrac{x}{8}\right)^{\frac{1}{3}} \\ &= 2\left(1 + \tfrac{x}{24} - \tfrac{x^2}{576} + \dots\right) \\ &= 2 + \tfrac{1}{12}x - \tfrac{1}{288}x^2 + \dots \end{aligned}

(b) Put $x = 1$ : $2 + \tfrac{1}{12} - \tfrac{1}{288} = \tfrac{599}{288} = 2.0799$ (4 d.p.).

(c) The expansion is only valid for $\left|\tfrac{x}{8}\right| < 1$ , that is $|x| < 8$ . $x = 19$ is outside this interval, so the series does not converge to $\sqrt[3]{27}$ .

Exam style

10 [8 marks]

f(x) = \frac{2 + x}{\sqrt{4 - x}}

Find the binomial expansion of $(4 - x)^{-\frac{1}{2}}$ up to the term in $x^2$ . [4]
State the values of $x$ for which it is valid. [1]
Hence find the expansion of $f(x)$ up to the term in $x^2$ . [3]

Show solution

(a) Take out $4^{-\frac{1}{2}} = \tfrac{1}{2}$ :

\begin{aligned} &\tfrac{1}{2}\left(1 - \tfrac{x}{4}\right)^{-\frac{1}{2}} \\ &= \tfrac{1}{2}\left(1 + \tfrac{x}{8} + \tfrac{3x^2}{128} + \dots\right) \\ &= \tfrac{1}{2} + \tfrac{1}{16}x + \tfrac{3}{256}x^2 + \dots \end{aligned}

The middle line uses $n = -\tfrac{1}{2}$ with $-\tfrac{x}{4}$ in place of $x$ : $\left(-\tfrac{1}{2}\right)\left(-\tfrac{x}{4}\right) = \tfrac{x}{8}$ and $\tfrac{3}{8}\left(-\tfrac{x}{4}\right)^2 = \tfrac{3x^2}{128}$ .

(b) $\left|\tfrac{x}{4}\right| < 1$ , that is $|x| < 4$ .

(c) Multiply by $2 + x$ and keep terms up to $x^2$ :

\begin{aligned} &1 + \tfrac{1}{8}x + \tfrac{3}{128}x^2 + \tfrac{1}{2}x + \tfrac{1}{16}x^2 \\ &= 1 + \tfrac{5}{8}x + \tfrac{11}{128}x^2 + \dots \end{aligned}

11 [9 marks]

f(x) = \frac{2 + 5x + x^2}{(1 - x)(1 + x)^2}

Express $f(x)$ in the form $\dfrac{A}{1 - x} + \dfrac{B}{1 + x} + \dfrac{C}{(1 + x)^2}$ . [4]
Hence show that, in the expansion of $f(x)$ in ascending powers of $x$ , the coefficient of $x^2$ is zero, and find the coefficient of $x^3$ . [4]
State the values of $x$ for which the expansion is valid. [1]

Show solution

(a) $2 + 5x + x^2 \equiv A(1 + x)^2 + B(1 - x)(1 + x) + C(1 - x)$ .

Put $x = 1$ : $8 = 4A$ , so $A = 2$ .
Put $x = -1$ : $-2 = 2C$ , so $C = -1$ .
Compare $x^2$ terms: $1 = A - B$ , so $B = 1$ .

f(x) = \frac{2}{1 - x} + \frac{1}{1 + x} - \frac{1}{(1 + x)^2}

(b) Expand each part:

\begin{aligned} \frac{2}{1 - x} &= 2 + 2x + 2x^2 + 2x^3 \\ \frac{1}{1 + x} &= 1 - x + x^2 - x^3 \\ -\frac{1}{(1 + x)^2} &= -1 + 2x - 3x^2 + 4x^3 \end{aligned}

The $x^2$ terms add to $2 + 1 - 3 = 0$ , and the $x^3$ terms to $2 - 1 + 4 = 5$ . So $f(x) = 2 + 3x + 5x^3 + \dots$ , and the coefficient of $x^3$ is $5$ .

(c) All three expansions are valid for $|x| < 1$ .

12 [9 marks]

Expand $\sqrt{1 - 8x}$ up to the term in $x^3$ . [3]
State the values of $x$ for which the expansion is valid. [1]
By substituting $x = \tfrac{1}{100}$ , show that $\sqrt{23} \approx 4.79584$ . [3]
Find the percentage error in this approximation, to 1 significant figure. [2]

Show solution

(a) Use $\sqrt{1 + x} = 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2 + \tfrac{1}{16}x^3 - \dots$ with $-8x$ in place of $x$ :

\begin{aligned} &1 - 4x - \tfrac{1}{8}(64x^2) + \tfrac{1}{16}(-512x^3) \\ &= 1 - 4x - 8x^2 - 32x^3 - \dots \end{aligned}

(b) $|8x| < 1$ , that is $|x| < \tfrac{1}{8}$ .

(c) When $x = \tfrac{1}{100}$ , the left side is $\sqrt{0.92} = \sqrt{\tfrac{4 \times 23}{100}} = \tfrac{1}{5}\sqrt{23}$ . The series gives

\begin{aligned} &1 - 0.04 - 0.0008 - 0.000032 \\ &= 0.959168 \end{aligned}

so $\sqrt{23} \approx 5 \times 0.959168 = 4.79584$ .

(d) $\sqrt{23} = 4.7958315\ldots$ , so the percentage error is

\begin{aligned} &\frac{4.79584 - 4.7958315\ldots}{4.7958315\ldots} \times 100 \\ &= 0.0002\% \end{aligned}

to 1 significant figure.