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Binomial Expansion Explained: A-Level Maths (9709)

Binomial expansion for Cambridge 9709: the formula for positive whole powers, finding one term, and the extended expansion for negative and fractional powers.

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Binomial expansion lets you expand a bracket such as (a+b)^n without multiplying it out repeatedly. For a positive whole-number power you use the binomial theorem, which gives a finite number of terms. For a negative or fractional power (Pure Mathematics 3) you use the extended expansion, which gives an infinite series that is only valid when |x| is small enough.

Positive whole-number powers (Pure 1)

In 9709 Pure Mathematics 1 you expand (a+b)^n where n is a positive integer:

$$(a+b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n$$

The symbol \binom{n}{r} is read "n choose r" and equals \frac{n!}{r!\,(n-r)!}. On a scientific calculator it is the nCr key. The syllabus expects you to know this notation.

Look at the pattern. The power of a falls by one each term while the power of b rises by one, and the two powers always add up to n. There are n+1 terms.

The binomial series for both versions appears in the list of formulae supplied in the exam, so you do not need to memorise the layout. What you do need is fluency in using it, which is where marks are won or lost.

Finding one term

Most questions ask for a single coefficient, such as "the coefficient of x^3". Expanding everything wastes time. Use the general term:

$$\text{term in } b^r = \binom{n}{r}\,a^{n-r}\,b^r$$

Worked example. Find the coefficient of x^3 in (1 + 2x)^{10}.

Here a = 1, b = 2x, n = 10. The x^3 term has r = 3:

$$\binom{10}{3}(1)^{7}(2x)^{3} = 120 \times 8x^3 = 960x^3$$

The coefficient is 960. The steps that go wrong are forgetting to cube the whole of 2x, so keeping the bracket around it protects you.

When both terms contain x

Some brackets have x in both parts, for example \left(3x - \frac{2}{x}\right)^6. The power of x depends on r, so you set up the general term first and simplify the powers of x before choosing r.

Worked example. Find the term independent of x in \left(3x - \frac{2}{x}\right)^6.

The general term is

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

For a term independent of x the power must be zero, so 6 - 2r = 0 and r = 3. Then

$$\binom{6}{3}\times 3^{3}\times(-2)^{3} = 20 \times 27 \times (-8) = -4320$$

The constant term is -4320. The negative sign comes from (-2)^3, which is easy to drop if you rush.

The extended expansion (Pure 3)

When n is negative or a fraction, the \binom{n}{r} notation no longer applies, so you use the series for (1+x)^n:

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

This series never stops, and it is valid only when |x| < 1.

Worked example. Expand (1+x)^{1/2} up to the term in x^2.

With n = \tfrac{1}{2} the terms are 1, then \tfrac{1}{2}x, then

$$\frac{\tfrac{1}{2}\left(\tfrac{1}{2}-1\right)}{2!}x^2 = \frac{\tfrac{1}{2}\times\left(-\tfrac{1}{2}\right)}{2}x^2 = -\tfrac{1}{8}x^2$$

So (1+x)^{1/2} \approx 1 + \tfrac{1}{2}x - \tfrac{1}{8}x^2, valid for |x| < 1.

Worked example with a negative power. Expand (1 - 3x)^{-1} up to x^2. Replace x by -3x and use n = -1:

$$1 + (-1)(-3x) + \frac{(-1)(-2)}{2!}(-3x)^2 = 1 + 3x + 9x^2$$

The condition is |-3x| < 1, which gives |x| < \tfrac{1}{3}. Notice that the validity range changes when the bracket is 1 + ax rather than 1 + x.

Getting the bracket into the right form

The standard series only works when the bracket starts with 1. For (4+x)^{1/2}, factor out the constant first:

$$(4+x)^{1/2} = \left(4\left(1 + \tfrac{x}{4}\right)\right)^{1/2} = 2\left(1 + \tfrac{x}{4}\right)^{1/2}$$

Expand \left(1 + \tfrac{x}{4}\right)^{1/2} using the series with x replaced by \tfrac{x}{4}: this gives 1 + \tfrac{x}{8} - \tfrac{x^2}{128}. Multiply by 2:

$$(4+x)^{1/2} \approx 2 + \tfrac{x}{4} - \tfrac{x^2}{64}$$

The condition is \left|\tfrac{x}{4}\right| < 1, so |x| < 4. Two things catch people out here: forgetting to multiply every term by the 2 you factored out, and quoting the condition for \tfrac{x}{4} instead of for x.

The two versions side by side

Positive integer power (Pure 1)Extended power (Pure 3)
Value of nPositive whole numberAny rational number
Number of termsFinite (n+1 terms)Infinite series
Uses \binom{n}{r}YesNo, uses the fraction form
Validity conditionNone needed\lvert x\rvert < 1 for (1+x)^n
Bracket must start with 1NoYes, factor out first

Spotting which form the question wants

Read the power first. If it is a positive whole number, you are in finite binomial territory and the nCr notation applies. If it is negative or a fraction, the bracket must be turned into the form (1 + \dots) before you do anything else, and a validity condition will almost certainly be part of the marks. Then read the last line of the question. "Find the coefficient of x^4" means use the general term and stop. "Expand up to and including the term in x^3" means write out each term in order. "Hence find an approximation" means the expansion is a tool for a later step, so keep your working tidy enough to reuse.

Checking the validity condition

Students often get the series right and lose the final mark on the range of x. Write the condition as soon as you decide what replaces x in the standard series. If the bracket is 1 + \tfrac{x}{4}, the thing being compared with 1 is \tfrac{x}{4}. If it is 1 - 3x, the sign does not matter because of the modulus, so you compare 3x with 1. Then solve the inequality for x itself, and check that you have written it as a range, for example -4 < x < 4 or |x| < 4.

Sign and bracket errors to watch for

Most lost marks in this topic are not conceptual. They come from handling brackets carelessly. When a term is negative, put it in brackets before you square or cube it, so (-3x)^2 is 9x^2 and not -9x^2. When the coefficient of x is not 1, substitute it into the series with brackets around it. When you factor out a constant, raise that constant to the same power as the bracket, so factoring 4 out of a bracket with power \tfrac{1}{2} gives a factor of 2, and with power -1 gives a factor of \tfrac{1}{4}.

Common mistakes

  • Forgetting to raise the coefficient inside the bracket to the power: (2x)^3 = 8x^3, not 2x^3.
  • Losing a minus sign when the bracket has subtraction, as in (1 - 3x)^n.
  • Stating the validity condition for the factored bracket without turning it back into a condition on x.
  • Mixing up "coefficient" and "term". The coefficient of x^3 above is 960, not 960x^3.

If you are working on exam technique more widely, how to answer proof questions in 9709 is a useful companion, because "show that" binomial questions need the same step-by-step discipline.

Practise binomial questions on past papers

Binomial questions reward accuracy over cleverness, so do plenty and check each line against a mark scheme. Choose a paper from the A Level Mathematics past papers, write your working by hand and have it marked against the official mark scheme, with follow-through marking applied. Reading and downloading papers is free, and a free account adds AI marking. The 2023 and 2024 papers are marked first, so check which years are available.

Frequently asked questions

What is the difference between the Pure 1 and Pure 3 binomial expansion?
In Pure Mathematics 1 the power is a positive whole number, so the expansion is finite. In Pure Mathematics 3 the power can be any rational number, including negative and fractional values, which gives an infinite series that is only valid for a limited range of x.
How do I find a single term without expanding everything?
Use the general term. In the expansion of a plus b to the power n, the term containing b to the power r is nCr times a to the power n minus r times b to the power r. Work out which r gives the power of x you want, then evaluate that one term.
Why does the extended expansion need a validity condition?
With a negative or fractional power the series never ends, so it only gives the true value when x is small enough. For one plus x to a rational power the condition is that the modulus of x is less than 1. If the bracket is one plus a x, the condition becomes the modulus of a x less than 1.
Are the binomial formulae given in the exam?
The list of formulae supplied in 9709 exams includes the binomial series for both a positive whole number n and a rational n. You still need to know how to use them, so practise with the list open.