Binomial Series for Any Rational Power

Expand (1 + x)ⁿ when n is negative or a fraction, state when the series is valid, rewrite brackets into that form, and use the series to estimate values.

In the exam for Edexcel Further Pure Maths 4PM1 · 19 real past-paper questions below

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A series that never stops

You can expand $(1 + x)^{5}$ . But what about $\frac{1}{1 + x}$ , which is $(1 + x)^{- 1}$ , or $1 + x$ , which is $(1 + x)^{\frac{1}{2}}$ ? The same series works, with one big change.

Formula

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

$n$ — any rational power: negative, or a fraction.
$x$ — must be small: $∣ x ∣ < 1$ when $n$ is not a positive whole number.

For a whole number $n$ , the factors $n$ , $n - 1$ , $n - 2$ , … reach 0, and the series stops. For $n = - 1$ or $n = \frac{1}{2}$ they never reach 0. So the series goes on for ever.

Worked example

1
Expand $(1 + x)^{- 1}$ up to the term in $x^{3}$ .
2
$1 + (- 1) x$ — $n = - 1$ .
3
$\frac{( - 1 ) ( - 2 )}{2} x^{2} = x^{2}$ — $n - 1 = - 2$ . Two minus signs make a plus.
4
$\frac{( - 1 ) ( - 2 ) ( - 3 )}{6} x^{3} = - x^{3}$ — $n - 2 = - 3$ . Three minus signs make a minus.
5
$1 - x + x^{2} - x^{3} + \dots$ — Check with $x = 0.1$ : $1 - 0.1 + 0.01 - 0.001 = 0.909$ , and $\frac{1}{1.1} = 0.90909 \dots$

Now try $x = 2$ : $1 - 2 + 4 - 8 + \dots$ The terms grow, and the sum never settles near $\frac{1}{3}$ . The series only works when $x$ is small.

Rule

The series is on the 4PM1 formula sheet, with $∣ x ∣ < 1$ . For your own expansion you must work out and state the validity yourself. Questions often give a mark just for it.

Key idea

Isaac Newton discovered this series for fractional and negative powers in about 1665, when he was in his early twenties.

Key idea

Whole-number power: the series stops. Any other power: it never stops, and it only works for small $x$ .

Fractional powers

Most questions put something like $2 x$ or $- \frac{x}{3}$ in place of $x$ . Replace $x$ by the whole thing, in brackets, everywhere.

Worked example

1
Expand $(1 - 2 x)^{\frac{1}{2}}$ up to the term in $x^{3}$ , and state when it is valid.
2
$n = \frac{1}{2}, x \to (- 2 x)$ — Name the parts. The minus sign goes inside the bracket.
3
$1 + \frac{1}{2} (- 2 x) = 1 - x$ — First two terms.
4
$\frac{\frac{1}{2} ( - \frac{1}{2} )}{2} (- 2 x)^{2} = - \frac{1}{8} \times 4 x^{2} = - \frac{1}{2} x^{2}$ — $n - 1 = - \frac{1}{2}$ . Square all of $- 2 x$ .
5
$\frac{\frac{1}{2} ( - \frac{1}{2} ) ( - \frac{3}{2} )}{6} (- 2 x)^{3} = \frac{1}{16} \times (- 8 x^{3})$ $= - \frac{1}{2} x^{3}$ — $n - 2 = - \frac{3}{2}$ . Cube all of $- 2 x$ .
6
$1 - x - \frac{1}{2} x^{2} - \frac{1}{2} x^{3}$ — Give each coefficient as a fraction in its lowest terms.
7
$∣ - 2 x ∣ < 1 \Rightarrow ∣ x ∣ < \frac{1}{2}$ — Valid when the thing replacing $x$ is between −1 and 1.

Mark schemes give a method mark for the correct structure, even before you simplify. Brackets must be there: a missing bracket loses the mark unless you recover it.

Common mistake

Writing $- 2 x^{2}$ instead of $(- 2 x)^{2} = 4 x^{2}$ . Always keep the bracket round the whole replacement.

Common mistake

Giving the validity as $∣ x ∣ < 1$ . It is the whole replacement that must be smaller than 1, so here $∣ x ∣ < \frac{1}{2}$ .

Key idea

Replace $x$ by the whole term, in brackets. Validity: that whole term between −1 and 1.

Negative powers and roots

First rewrite the expression as a bracket to a power. A bracket on the bottom has a negative power. A root is a fraction power.

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

Worked example

1
Expand $\frac{1}{( 1 + 3 x ) ^{2}}$ up to the term in $x^{3}$ , and state when it is valid.
2
$(1 + 3 x)^{- 2}, n = - 2$ — Rewrite with a negative power.
3
$1 + (- 2) (3 x) = 1 - 6 x$ — First two terms.
4
$\frac{( - 2 ) ( - 3 )}{2} (3 x)^{2} = 3 \times 9 x^{2} = 27 x^{2}$ — $n - 1 = - 3$ . $(3 x)^{2} = 9 x^{2}$ .
5
$\frac{( - 2 ) ( - 3 ) ( - 4 )}{6} (3 x)^{3} = - 4 \times 27 x^{3}$ $= - 108 x^{3}$ — $n - 2 = - 4$ . $(3 x)^{3} = 27 x^{3}$ .
6
$1 - 6 x + 27 x^{2} - 108 x^{3}, ∣ x ∣ < \frac{1}{3}$ — Valid when $∣3 x ∣ < 1$ .

Common mistake

Getting $n - 1$ wrong for a negative $n$ . When $n = - 2$ , $n - 1 = - 3$ , not −1. Moving down from a negative number makes it more negative.

Key idea

Rewrite as a power first. Then the signs of $n$ , $n - 1$ , $n - 2$ decide the signs of the terms.

When the bracket does not start with 1

The series needs $(1 + \dots)$ . If the bracket starts with another number, take that number out first. Remember it is raised to the power too.

Formula

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

$a$ — the number at the front of the bracket. Divide every term inside by it, and multiply by $a^{n}$ outside.

Worked example

1
Expand $\frac{1}{( 2 - x ) ^{2}}$ up to the term in $x^{3}$ , and state when it is valid.
2
$(2 - x)^{- 2} = 2^{- 2} (1 - \frac{x}{2})^{- 2}$ $= \frac{1}{4} (1 - \frac{x}{2})^{- 2}$ — Take out the 2. $2^{- 2} = \frac{1}{4}$ .
3
$1 + (- 2) (- \frac{x}{2}) = 1 + x$ — Expand the bracket. First two terms.
4
$\frac{( - 2 ) ( - 3 )}{2} (- \frac{x}{2})^{2} = 3 \times \frac{x ^{2}}{4} = \frac{3}{4} x^{2}$ — Third term.
5
$\frac{( - 2 ) ( - 3 ) ( - 4 )}{6} (- \frac{x}{2})^{3} = - 4 \times (- \frac{x ^{3}}{8})$ $= \frac{1}{2} x^{3}$ — Fourth term.
6
$\frac{1}{4} (1 + x + \frac{3}{4} x^{2} + \frac{1}{2} x^{3})$ $= \frac{1}{4} + \frac{1}{4} x + \frac{3}{16} x^{2} + \frac{1}{8} x^{3}$ — Multiply every term by $\frac{1}{4}$ .
7
$\frac{x}{2} < 1 \Rightarrow ∣ x ∣ < 2$ — Validity.

Common mistake

Taking out 2 instead of $2^{- 2}$ . The number you take out keeps the power: here $\frac{1}{4}$ , not 2.

Common mistake

Forgetting to multiply every term by the number outside at the end.

Key idea

Take out $a^{n}$ , expand $(1 + \frac{b}{a} x)^{n}$ , multiply back. Valid for $∣ x ∣ < \frac{a}{b}$ , ignoring signs.

Estimating with the series

A few terms of the series give a very good estimate, as long as $x$ is small. Choose $x$ so the bracket becomes the number you need.

Worked example

1
Use $(1 - 2 x)^{\frac{1}{2}} \approx 1 - x - \frac{1}{2} x^{2} - \frac{1}{2} x^{3}$ to estimate $0.96$ , then $6$ .
2
$1 - 2 x = 0.96 \Rightarrow x = 0.02$ — Choose $x$ . Check it is valid: $0.02 < \frac{1}{2}$ .
3
$1 - 0.02 - \frac{1}{2} (0.0004) - \frac{1}{2} (0.000008)$ — Put $x = 0.02$ into every term.
4
$0.96 \approx 0.979796$ — The true value is 0.9797959…, so the estimate is excellent.
5
$0.96 = \frac{96}{100} = \frac{4 6}{10} = 0.4 6$ — Link it to $6$ : $96 = 46$ .
6
$6 \approx 0.979796 \div 0.4 = 2.44949$ — To 5 decimal places.

Marks: one for the right value of $x$ , one for putting it into the series, one for the answer to the accuracy asked. A calculator value with no series earns nothing.

Some questions ask you to estimate an integral. Integrate the series term by term, instead of the expression itself.

Worked example

1
Use $(1 + 3 x)^{- 2} \approx 1 - 6 x + 27 x^{2} - 108 x^{3}$ to estimate $\int_{0}^{0.05} \frac{1}{( 1 + 3 x ) ^{2}} d x$ to 4 decimal places.
2
$\int_{0}^{0.05} (1 - 6 x + 27 x^{2} - 108 x^{3}) d x$ — Replace the expression by its series. 0.05 is inside the valid range, |x| < 1/3.
3
$= [x - 3 x^{2} + 9 x^{3} - 27 x^{4}]_{0}^{0.05}$ — Integrate each term: add 1 to the power, then divide by the new power.
4
$= 0.05 - 0.0075 + 0.001125 - 0.00016875$ — Put in 0.05. At 0 every term is 0.
5
$= 0.04345625 \approx 0.0435$ — To 4 decimal places.

Common mistake

Choosing an $x$ outside the validity range, such as $x = 0.6$ here. The series then gives nonsense.

Key idea

Make the bracket equal the number you want, check $x$ is valid and small, then substitute.

Products and unknowns

Exam questions often multiply a series by another bracket, or hide $a$ and $n$ in it. Use the same skills as for whole-number powers.

Worked example

1
Expand $\frac{2 + x}{( 1 + 3 x ) ^{2}}$ up to the term in $x^{2}$ .
2
$(2 + x) (1 - 6 x + 27 x^{2} + \dots)$ — Use the series for $(1 + 3 x)^{- 2}$ from before.
3
$2 - 12 x + 54 x^{2}$ — Multiply the series by 2. Stop at $x^{2}$ .
4
$x - 6 x^{2}$ — Multiply by $x$ . Stop at $x^{2}$ .
5
$2 - 11 x + 48 x^{2}$ — Add like terms. Valid for $∣ x ∣ < \frac{1}{3}$ , like the series.

Worked example

1
In the expansion of $(1 + a x)^{n}$ , the coefficient of $x$ is −6 and the coefficient of $x^{2}$ is 27. Find $a$ and $n$ .
2
$na = - 6$ — The $x$ term.
3
$\frac{n ( n - 1 )}{2} a^{2} = 27 \Rightarrow \frac{( n - 1 ) a \times na}{2} = 27$ — The $x^{2}$ term. Split $n (n - 1) a^{2}$ .
4
$(n - 1) a \times (- 3) = 27 \Rightarrow na - a = - 9$ — Use $na = - 6$ , then divide by −3.
5
$- 6 - a = - 9, a = 3, n = - 2$ — So the bracket is $(1 + 3 x)^{- 2}$ , the series from before.

Common mistake

Keeping terms beyond the power asked for. Ignore anything above $x^{2}$ if the question stops there.

Key idea

A product: multiply each term of the short bracket by the series, then collect. Unknowns: form equations from the coefficients.

Which method? Read the question

The wording tells you which step to use.

You see	You do
A fraction or root	Rewrite as a bracket to a power
Bracket starts with a number other than 1	Take out $a^{n}$ first
“Expand up to the term in $x^{3}$ ”	Replace $x$ by the whole term, in brackets; four terms
“State the values of $x$ for which it is valid”	Whole replaced term between −1 and 1
“Use your expansion to estimate”	Choose $x$ to make the bracket the number; check validity
Series times another bracket	Multiply term by term; stop at the power asked
Estimate an integral of the expression	Integrate the series term by term, then put in the limits

Common mistake

Leaving coefficients as decimals. Give exact fractions in lowest terms.

Key idea

Rewrite, expand with brackets, state validity. These three habits win most of the marks.

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