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Cambridge International AS & A Level · Topic 1.6
A-Level Maths 9709 / Topic 1.6 / AS & A Level

Every bracket, expanded.

Multiplying out \((2 + x)^5\) by hand takes a page; the binomial theorem does it in one line. Learn \(\binom{n}{r}\), the pattern of powers, and how to pick out a single term, such as a coefficient or the term independent of \(x\).

Topic 1.6All series lessons →
The key idea

\((a + b)^n = \displaystyle\sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^{r}\), where \(\binom{n}{r} = \dfrac{n!}{r!\,(n-r)!}\). The general term is \(\binom{n}{r}a^{n-r}b^{r}\): the powers of \(a\) and \(b\) always add up to \(n\).

1 11 121 1331 14641 15101051 1615201561 n = 0n = 1n = 2n = 3 n = 4n = 5n = 6 (a + b)⁵
Fig. 1: Pascal's triangle. Each entry is the sum of the two above it (4 + 6 = 10). Row n = 5 gives the coefficients of (a + b)⁵, and its entries are 5C0, 5C1, ..., 5C5
Section 01

Pascal's triangle and nCr.

Multiply out \((a + b)^2\), \((a + b)^3\), \((a + b)^4\) and the coefficients build Pascal's triangle (Fig. 1). For large \(n\) you need a formula for the entries: \(\binom{n}{r}\), read as \(n\) choose \(r\), and written \({}^{n}C_{r}\) on most calculators.

Worked example Coefficients without a calculator
Find the values of \(\binom{6}{2}\) and \(\binom{7}{3}\).
  1. \(\binom{6}{2} = \dfrac{6!}{2!\,4!} = \dfrac{6 \times 5}{2 \times 1} = 15\) (the \(4!\) cancels).
  2. \(\binom{7}{3} = \dfrac{7!}{3!\,4!} = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1} = \dfrac{210}{6} = 35\).
Answer: \(\binom{6}{2} = 15\), \(\binom{7}{3} = 35\)
Examiner trap

\(\binom{5}{2}\) is not \(\dfrac{5!}{2!} = 60\). Both factorials go underneath: \(\dfrac{5!}{2!\,3!} = 10\).

Section 02

Expanding (a + b)n.

Write the coefficients from \(\binom{n}{0}\) to \(\binom{n}{n}\); let the power of the first term fall from \(n\) to 0 while the power of the second rises from 0 to \(n\). Put brackets round each term, especially when it has a number or a minus sign in it.

Worked example A full expansion
Expand \((2 + x)^5\) fully.
  1. Row 5: \(1, 5, 10, 10, 5, 1\). Powers of 2 fall from \(2^5\); powers of \(x\) rise from \(x^0\).
  2. \(1(2^5) + 5(2^4)x + 10(2^3)x^2 + 10(2^2)x^3 + 5(2)x^4 + x^5\)
  3. \(= 32 + 5 \times 16x + 10 \times 8x^2 + 10 \times 4x^3 + 10x^4 + x^5\)
Answer: \(32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5\)
Worked example The first three terms
Find the first three terms, in ascending powers of \(x\), of the expansion of \((1 - 3x)^7\).
  1. \(1^7 + \binom{7}{1}(-3x) + \binom{7}{2}(-3x)^2 + \cdots\)
  2. \(\binom{7}{1} = 7\), so the second term is \(7 \times (-3x) = -21x\).
  3. \(\binom{7}{2} = 21\) and \((-3x)^2 = 9x^2\), so the third term is \(189x^2\).
Answer: \(1 - 21x + 189x^2\)
Your turn
Find the first three terms, in ascending powers of \(x\), of \((3 + 2x)^5\).
Examiner trap

Bracket the whole second term before squaring: \((-3x)^2 = 9x^2\), not \(-3x^2\) and not \(-9x^2\).

Stage 1 · Learn

Check your understanding

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Find the value of \(\binom{5}{2}\).

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What is the coefficient of \(x^2\) in the expansion of \((1 + x)^6\)?

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How many terms are there in the expansion of \((a + b)^7\)?

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What is the third term, in ascending powers of \(x\), of \((1 + 2x)^5\)?

Section 03

Picking out one term.

You rarely need the whole expansion. Write the general term \(\binom{n}{r}a^{n-r}b^{r}\), simplify the power of \(x\), and choose \(r\) to give the power you want.

Worked example A single coefficient
Find the coefficient of \(x^3\) in the expansion of \((3 - 2x)^6\).
  1. The \(x^3\) term has \(r = 3\): \(\binom{6}{3}(3)^{3}(-2x)^{3}\).
  2. \(\binom{6}{3} = 20\), \(3^3 = 27\), \((-2)^3 = -8\).
  3. \(20 \times 27 \times (-8) = -4320\).
Answer: \(-4320\)
Worked example The term independent of x
Find the term independent of \(x\) in the expansion of \(\left(x^2 + \dfrac{2}{x}\right)^6\).
  1. General term: \(\binom{6}{r}(x^2)^{6-r}\left(\dfrac{2}{x}\right)^{r} = \binom{6}{r}2^{r}x^{12 - 2r}x^{-r} = \binom{6}{r}2^{r}x^{12-3r}\).
  2. Independent of \(x\) means power 0: \(12 - 3r = 0\), so \(r = 4\).
  3. \(\binom{6}{4}2^4 = 15 \times 16\).
Answer: \(240\)
Examiner trap

The \(x^3\) term is the FOURTH term (\(r = 3\)), because counting starts at \(r = 0\). Using \(\binom{6}{4}\) or \(3^4\) here is the most common slip.

Section 04

Products and unknowns.

For a product such as \((1 + 2x)(1 - x)^6\), expand only as far as you need, then collect every pair of terms that multiplies to the power you want. If a coefficient contains an unknown, set it equal to the value given and solve.

Worked example Coefficient in a product
Find the coefficient of \(x^2\) in the expansion of \((1 + 2x)(1 - x)^6\).
  1. \((1 - x)^6 = 1 - 6x + 15x^2 + \cdots\)
  2. \(x^2\) terms come from \(1 \times 15x^2\) and \(2x \times (-6x)\).
  3. \(15 - 12 = 3\).
Answer: \(3\)
Worked example Finding an unknown
The coefficient of \(x^2\) in the expansion of \((1 + ax)^5\) is 40. Find the possible values of \(a\).
  1. The \(x^2\) term is \(\binom{5}{2}(ax)^2 = 10a^2x^2\).
  2. \(10a^2 = 40\), so \(a^2 = 4\).
Answer: \(a = 2\) or \(a = -2\)
Your turn
Find the coefficient of \(x^2\) in the expansion of \((2 + x)^4(1 - 3x)\).
Examiner trap

Collect ALL the pairs. In \((1 + 2x)(1 - x)^6\) the answer is not just 15: the \(2x\) multiplies the \(x\) term to make a second \(x^2\) term.

Stage 2 · Exam

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Find the coefficient of \(x^3\) in the expansion of \((2 - x)^7\).

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Find the term independent of \(x\) in the expansion of \(\left(x + \dfrac{3}{x^2}\right)^6\).

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The coefficient of \(x^2\) in the expansion of \((1 + ax)^6\) is 60, and \(a > 0\). Find \(a\).

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Find the coefficient of \(x^2\) in the expansion of \((1 + 2x)(1 - 3x)^5\).

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Binomial expansions, expanded.

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Six original Cambridge-style questions
Binomial coefficients, full and partial expansions, terms independent of \(x\), products, and compound interest.