P2.4 The binomial expansion
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Pure Mathematics 2 (AS), specification section 4.5: expanding \((a + bx)^n\) for a positive integer \(n\). Negative and fractional powers follow in P4.
What this page covers
7 points
- Factorials \(n!\), the notation \(\binom{n}{r}\) and its meaning as a number of choices
- Pascal's triangle and its addition rule
- Expanding \((a + bx)^n\) fully, or its first few terms in ascending powers of \(x\)
- Finding one particular term or coefficient with the general term
- Finding an unknown constant, or the power \(n\) itself, from given coefficients
- Using the first few terms to approximate a value such as \(0.995^{10}\)
- Coefficients in the product of a short bracket and a binomial expansion
In one minute
6 points
- The binomial expansion writes \((a + b)^n\) as \(n + 1\) terms, \(a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n\), with the powers of \(a\) falling as the powers of \(b\) rise.
- \(n! = n \times (n - 1) \times \dots \times 2 \times 1\), with \(0! = 1\); \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\) counts the ways to choose \(r\) objects from \(n\).
- Row \(n\) of Pascal's triangle lists \(\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}\); each entry is the sum of the two above it.
- General term: \(\binom{n}{r}a^{n-r}(bx)^r\) is the term in \(x^r\). Keep the bracket: the sign and the number inside are both raised to the power.
- Given coefficients turn into equations in the unknowns; solve them like any other simultaneous equations.
- For a small \(x\) the terms shrink fast, so the first few terms give an accurate approximation.
Contents
14 sections
- What do \(n!\) and \(\binom{n}{r}\) count?
- How does Pascal's triangle give the coefficients?
- How do you expand \((a + bx)^n\) in ascending powers of \(x\)?
- How do you find one particular term or coefficient?
- How do you find an unknown from given coefficients?
- How do you use an expansion to approximate a value?
- How do you find a coefficient in a product of two brackets?
- How do longer binomial questions fit together?
- Exam tips for P2.4
- Common misconceptions
- How this topic connects
- Key formulae
- Check your understanding
- Specification checklist
What do \(n!\) and \(\binom{n}{r}\) count?
This page rests on one formula, the general term \(\binom{n}{r}a^{n-r}b^r\), and on three habits that make it reliable: brackets round every power of the second term, signs carried through, and a quick check on the size of each coefficient. Expansions, single coefficients, unknowns and approximations then all follow the same pattern. First, the numbers that appear as coefficients.
- \(n!\) ("\(n\) factorial") is the number of ways to arrange \(n\) different objects in a line: for example \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). By definition \(0! = 1\), which keeps the formulas working when \(r = 0\) or \(r = n\).
- \(\binom{n}{r}\) (also written \({}^nC_r\), read "\(n\) choose \(r\)") is the number of ways to choose \(r\) objects from \(n\) when order does not matter: \(\binom{5}{2} = 10\).
- Symmetry: choosing \(r\) to take is the same as choosing \(n - r\) to leave, so \(\binom{n}{r} = \binom{n}{n - r}\). For example \(\binom{10}{7} = \binom{10}{3} = 120\).
- Why choices give coefficients: multiplying out \((a + b)^n\) means picking \(a\) or \(b\) from each of the \(n\) brackets. A term \(a^{n-r}b^r\) arises once for every way of choosing which \(r\) brackets supply the \(b\), so its coefficient is \(\binom{n}{r}\).
(a) Use the definition in terms of factorials to evaluate \(\binom{7}{3}\). (b) Show that \(\binom{n}{2} = \frac12 n(n - 1)\), and hence find the positive integer \(n\) for which \(\binom{n}{2} = 66\). (c) Without evaluating each term separately, find \(\binom{7}{3} + \binom{7}{4}\).
(a) Write out the factorials and cancel: \(\binom{7}{3} = \frac{7!}{3!\,4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} =\) 35.
(b) Cancel \((n - 2)!\) from \(n!\): \(\binom{n}{2} = \frac{n!}{2!\,(n - 2)!} = \frac{n(n - 1)}{2}\). Then \(\frac12 n(n - 1) = 66\), so \(n^2 - n - 132 = 0\), \((n - 12)(n + 11) = 0\), giving \(n = 12\). (\(n = -11\) is rejected: \(n\) counts objects, so it must be a positive integer.)
(c) Two neighbours in row 7 add to the entry below them in row 8: \(\binom{7}{3} + \binom{7}{4} = \binom{8}{4} =\) 70.
How does Pascal's triangle give the coefficients?
- Write 1 at the top (row 0). Each later row starts and ends with 1, and every other entry is the sum of the two entries above it.
- Row \(n\) lists \(\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}\): the coefficients of \((a + b)^n\). The addition rule is the identity \(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\).
- Why the rule works: in \((a + b)^{n+1} = (a + b)(a + b)^n\), each term of the new row collects one contribution from the \(a\) and one from the \(b\), which are the two coefficients above it.
- The triangle is quickest for small powers (up to about 7). For larger \(n\), or for a single term, use \(\binom{n}{r}\) on a calculator.
Expand \((2 + x)^5\) fully, simplifying each term.
Coefficients from row 5; powers of 2 fall from 5 to 0 while powers of \(x\) rise from 0 to 5:
\(1(2^5) + 5(2^4)x + 10(2^3)x^2 + 10(2^2)x^3 + 5(2)x^4 + x^5\)
Simplify each term: \((2 + x)^5 = 32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5\)
Check: put \(x = 1\). The left side is \(3^5 = 243\) and the coefficients add to 243. ✓
How do you expand \((a + bx)^n\) in ascending powers of \(x\)?
- \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\); \(n\) a positive integer; there are \(n + 1\) terms.
- In the formulae list? In the formulae booklet: learn how to apply it.
For \((a + bx)^n\), replace \(b\) by the whole second term \((bx)\), bracket and sign included.
- "Ascending powers of \(x\)" means the constant first, then \(x, x^2, \dots\). "The first four terms" therefore stops at \(x^3\).
- Write each term as coefficient × power of \(a\) × (second term)power, then simplify. Doing the bracketed power in one step, such as \((-2x)^3 = -8x^3\), avoids most sign errors.
- If the second term is negative the signs alternate; if the first term is not 1, its powers multiply every coefficient.
Find the first four terms, in ascending powers of \(x\), of the binomial expansion of \((3 - 2x)^6\), giving each term in its simplest form.
Here \(a = 3\) and the second term is \((-2x)\):
\(3^6 + \binom61(3^5)(-2x) + \binom62(3^4)(-2x)^2 + \binom63(3^3)(-2x)^3\)
\(= 729 + 6(243)(-2x) + 15(81)(4x^2) + 20(27)(-8x^3)\)
\(= 729 - 2916x + 4860x^2 - 4320x^3\)
Writing \(-2x^2\) for \((-2x)^2\). The bracket means the \(-2\) is squared too, giving \(4x^2\). Without it the third term above becomes \(-2430x^2\) instead of \(+4860x^2\): wrong in both sign and size.
How do you find one particular term or coefficient?
- The power of \(x\) tells you \(r\) directly. The term in \(x^r\) is the \((r + 1)\)th term, because the expansion starts with \(x^0\).
- Term and coefficient are different answers: the term includes the power of \(x\), the coefficient is just the number in front of it.
- You never need the earlier terms: go straight to the \(r\) you want.
In the binomial expansion of \(\left(2 + \frac{x}{4}\right)^9\), find (a) the coefficient of \(x^4\), (b) the coefficient of \(x^5\), giving each as an exact fraction.
(a) Term in \(x^4\), so \(r = 4\): \(\binom94 \times 2^5 \times \left(\frac{x}{4}\right)^4 = 126 \times 32 \times \frac{1}{256}x^4\). Coefficient \(= \frac{4032}{256} =\) \(\frac{63}{4}\).
(b) Term in \(x^5\), so \(r = 5\): \(\binom95 \times 2^4 \times \left(\frac{x}{4}\right)^5 = 126 \times 16 \times \frac{1}{1024}x^5\). Coefficient \(= \frac{2016}{1024} =\) \(\frac{63}{32}\).
Check by ratio: from one coefficient to the next, multiply by \(\frac{n - r}{r + 1} \times \frac{b}{a}\). Here \(\frac55 \times \frac{1/4}{2} = \frac18\), and \(\frac{63}{4} \times \frac18 = \frac{63}{32}\). ✓
How do you find an unknown from given coefficients?
- Write each coefficient you are told about in terms of the unknowns, using the general term, then set it equal to its given value.
- One unknown needs one equation. Two unknowns, such as \(n\) and \(k\) in \((1 + kx)^n\), need two coefficients.
- When \(n\) is unknown, write \(\binom{n}{2} = \frac{n(n - 1)}{2}\) and \(\binom{n}{3} = \frac{n(n - 1)(n - 2)}{6}\). Dividing one equation by the other often cancels the awkward factors.
- Use the conditions in the question to reject solutions: \(n\) must be a positive integer, and the question may say a constant is positive or non-zero.
In the binomial expansion of \((1 + kx)^n\), where \(n\) is a positive integer and \(k\) is a constant, the coefficient of \(x\) is 24 and the coefficient of \(x^2\) is 252. (a) Find \(n\) and \(k\). (b) Find the coefficient of \(x^3\).
(a) Coefficients in terms of \(n\) and \(k\): \(nk = 24\) and \(\frac{n(n - 1)}{2}k^2 = 252\).
Divide the second equation by the first: \(\frac{(n - 1)k}{2} = \frac{252}{24} = \frac{21}{2}\), so \((n - 1)k = 21\).
Subtract from \(nk = 24\): \(nk - (n - 1)k = k = 24 - 21\), so \(k = 3\), \(n = 8\).
(b) Term in \(x^3\): \(\binom83 \times 3^3 = 56 \times 27 =\) 1512.
Check: \((1 + 3x)^8 = 1 + 24x + 252x^2 + 1512x^3 + \dots\) ✓
"The coefficients of \(x^2\) and \(x^3\) are equal" is an equation too. In \((2 + ax)^6\) they are \(15 \times 2^4a^2 = 240a^2\) and \(20 \times 2^3a^3 = 160a^3\). Setting them equal and dividing by \(80a^2\) (allowed because \(a \ne 0\)) gives \(3 = 2a\), so \(a = \frac32\) and both coefficients are 540. Divide by the common factor rather than cancelling it silently.
How do you use an expansion to approximate a value?
- When \(x\) is small, each power \(x^r\) is much smaller than the one before, so the later terms hardly change the total. The first three or four terms give a good approximation.
- To approximate a number such as \(0.995^{10}\), match it to the bracket: \(1 - \frac{x}{4} = 0.995\) gives \(x = 0.02\). Then substitute that \(x\) into your terms.
- The approximation is good only near \(x = 0\). Away from it, the neglected terms are no longer small and the polynomial peels away from the true curve.
- To judge accuracy, compare with a calculator value, or look at the size of the first term you left out.
(a) Find the first four terms, in ascending powers of \(x\), of the binomial expansion of \(\left(1 - \frac{x}{4}\right)^{10}\). (b) Use your expansion, with a suitable value of \(x\), to find an approximate value of \(0.995^{10}\). Give your answer to 5 decimal places.
(a) General term with \(a = 1\) and second term \(-\frac{x}{4}\): \(1 + 10\left(-\frac{x}{4}\right) + 45\left(-\frac{x}{4}\right)^2 + 120\left(-\frac{x}{4}\right)^3\)
\(= 1 - \frac52 x + \frac{45}{16}x^2 - \frac{15}{8}x^3\)
(b) Choose \(x\): \(1 - \frac{x}{4} = 0.995\), so \(x = 0.02\). Then \(1 - 0.05 + 0.001125 - 0.000015 =\) 0.95111 (5 d.p.).
A calculator gives \(0.995^{10} = 0.951110130\dots\): the four-term estimate is out by about \(1.3 \times 10^{-7}\), a percentage error of about \(1.4 \times 10^{-5}\,\%\).
How do you find a coefficient in a product of two brackets?
- For \((p + qx)(a + bx)^n\), expand the binomial only as far as you need, then multiply by the short bracket.
- An \(x^r\) term can arise in more than one way: the constant \(p\) times the \(x^r\) term, plus \(qx\) times the \(x^{r-1}\) term. Add all the pairs whose powers sum to \(r\).
- To reach \(x^3\) in the product you need the binomial terms up to \(x^3\); terms beyond that can never contribute.
(a) Find the first four terms, in ascending powers of \(x\), of the binomial expansion of \((1 + 2x)^8\). (b) Hence find the coefficient of \(x^2\) and the coefficient of \(x^3\) in the expansion of \((3 - x)(1 + 2x)^8\).
(a) General term with second term \(2x\): \(1 + 8(2x) + 28(2x)^2 + 56(2x)^3 =\) \(1 + 16x + 112x^2 + 448x^3\).
(b) \(x^2\): \(3 \times 112x^2\) and \(-x \times 16x\), so the coefficient is \(3 \times 112 - 16 =\) 320.
\(x^3\): \(3 \times 448x^3\) and \(-x \times 112x^2\), so the coefficient is \(3 \times 448 - 112 =\) 1232.
"Hence" means use part (a): no new expansion is needed, only the right pairs of terms.
How do longer binomial questions fit together?
A typical longer question builds in layers: an expansion with an unknown, an equation from a given coefficient, a further coefficient, then a numerical use of the result. Keep the unknown in your terms until you have its value, then substitute once into simplified coefficients. For an approximation from a product, find the \(x\) that makes both brackets match the numbers you are given.
(a) Find the first four terms, in ascending powers of \(x\), of the binomial expansion of \((1 + px)^8\), where \(p\) is a constant. (b) In the expansion of \((2 - x)(1 + px)^8\), the coefficient of \(x^2\) is 10. Given that \(p > 0\), find \(p\). (c) Find the coefficient of \(x^3\) in the expansion of \((2 - x)(1 + px)^8\). (d) Use your answers, with a suitable value of \(x\), to estimate \(1.98 \times 1.01^8\), giving your answer to 6 decimal places.
(a) \((1 + px)^8 = 1 + 8px + 28p^2x^2 + 56p^3x^3 + \dots\)
(b) Pairs that give \(x^2\): \(2 \times 28p^2\) and \(-1 \times 8p\). So \(56p^2 - 8p = 10\), \(28p^2 - 4p - 5 = 0\), \((14p + 5)(2p - 1) = 0\). \(p > 0\), so \(p = \frac12\) (reject \(p = -\frac{5}{14}\)).
(c) Pairs that give \(x^3\): \(2 \times 56p^3\) and \(-1 \times 28p^2\): \(2 \times 56 \times \frac18 - 28 \times \frac14 = 14 - 7 =\) 7.
(d) Full product to \(x^3\) with \(p = \frac12\); the \(x\) coefficient is \(2 \times 4 - 1 = 7\): \((2 - x)\left(1 + \frac{x}{2}\right)^8 \approx 2 + 7x + 10x^2 + 7x^3\).
\(x = 0.02\) gives \(2 - x = 1.98\) and \(1 + \frac{x}{2} = 1.01\), so both brackets match: \(2 + 0.14 + 0.004 + 0.000056 =\) 2.144056.
A calculator gives \(1.98 \times 1.01^8 = 2.144056277\dots\), so the estimate is correct to 6 decimal places.
Same method, different numbers: if a question gives a negative or fractional power, the series never stops. That is the P4 binomial series, valid only for \(|x|\) small enough.
Exam tips for P2.4
One line of the form \(\binom{n}{r} \times a^{n-r} \times (\text{second term})^r\) for each term shows your method clearly, so a slip in the arithmetic afterwards costs as little as possible.
Read which one is asked for. "The term in \(x^3\)" wants \(-4320x^3\); "the coefficient of \(x^3\)" wants \(-4320\). Leave fractions exact unless a decimal is requested.
Show the equation that fixes \(x\) (for example \(1 - \frac{x}{4} = 0.995\)), then the substituted terms, then the answer to the accuracy asked for.
Substituting \(x = 1\) into a full expansion must give the bracket's value, for example \((2 + 1)^5 = 243\). For a partial expansion, check one coefficient again with the general term.
When \(n\) is unknown, write \(\binom{n}{2}\) and \(\binom{n}{3}\) as products in \(n\) straight away. Divide one coefficient equation by another before expanding: it usually removes the highest power of the unknown constant.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| \((a + b)^2 = a^2 + b^2\), and similarly for higher powers. | Multiplying out produces cross terms such as \(2ab\). | \((a + b)^n\) has \(n + 1\) terms with coefficients \(\binom{n}{0}, \dots, \binom{n}{n}\). |
| \((-2x)^3 = -2x^3\). | The power applies to everything in the bracket. | \((-2x)^3 = -8x^3\): cube the number and the sign as well. |
| The term in \(x^4\) uses \(r = 5\), because it is the fifth term. | The expansion starts with \(x^0\), so term numbers are one ahead of powers. | The term in \(x^r\) uses \(\binom{n}{r}\). |
| The coefficient of \(x\) in \((3 + 2x)^5\) is just \(\binom51 = 5\). | The first and second terms also contribute their powers. | \(5 \times 3^4 \times 2 = 810\): every coefficient includes \(a^{n-r}b^r\). |
| The coefficient of \(x^2\) in a product is the product of the two \(x^2\) coefficients. | \(x^2\) also comes from \(x \times x\) and from a constant times \(x^2\). | Add every pair of terms whose powers sum to 2. |
| The first few terms give a good approximation for any \(x\). | For larger \(x\) the neglected terms are not small. | Use a small \(x\), close to 0; compare with a calculator value if asked about accuracy. |
How this topic connects
P1 algebra and P2.3 sequences and series
Expanding brackets and solving quadratics, used here to find \(n\) or a constant from a coefficient, are P1 skills. The ratio of successive coefficients, and the idea that later terms of a series can become negligible, link to geometric series and their sums.
P4 the binomial series
\((1 + x)^n\) extends to negative and fractional \(n\) as an infinite series, valid for \(|x| < 1\), and is used for approximations and with partial fractions.
S2 the binomial distribution
The probability of \(r\) successes in \(n\) trials is \(\binom{n}{r}p^r(1 - p)^{n-r}\): one term of the expansion of \((q + p)^n\).
Key formulae
| Formula | Use | In the formulae list? |
|---|---|---|
| \(n! = n(n - 1)\cdots 2 \times 1\), \(0! = 1\) | factorials; cancelling in \(\binom{n}{r}\) | No — learn it |
| \(\binom{n}{r} = \frac{n!}{r!(n - r)!}\) | number of choices; binomial coefficients | Yes, with the binomial expansion |
| \(\binom{n}{r} = \binom{n}{n - r}\) | symmetry of each row | No — derive it |
| \(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\) | building Pascal's triangle | No — learn it |
| \((a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \dots + \binom{n}{r}a^{n-r}b^r + \dots + b^n\) | full or partial expansions, positive integer \(n\) | Yes |
| term in \(x^r\): \(\binom{n}{r}a^{n-r}(bx)^r\) | one particular term or coefficient | Yes, as the general term of the expansion |
| \(\binom{n}{2} = \frac{n(n - 1)}{2}\), \(\binom{n}{3} = \frac{n(n - 1)(n - 2)}{6}\) | equations when \(n\) is unknown | No — derive it |
Check your understanding
1. Why must \(0!\) equal 1 for the formula for \(\binom{n}{r}\) to work?
\(\binom{n}{0}\) counts the ways of choosing nothing, which is 1. The formula gives \(\frac{n!}{0!\,n!} = \frac{1}{0!}\), which equals 1 only if \(0! = 1\).
2. Find the coefficient of \(x^2\) in the expansion of \((1 - 3x)^5\).
\(\binom52 \times (-3)^2 = 10 \times 9\). The coefficient is 90.
3. Evaluate \(\binom{20}{17}\) without a calculator.
\(\binom{20}{17} = \binom{20}{3} = \frac{20 \times 19 \times 18}{6}\). 1140. Choosing 17 to take is the same as choosing 3 to leave.
4. You have the first three terms of \(\left(1 + \frac{x}{2}\right)^6\). What value of \(x\) would you use to estimate \(1.03^6\)?
\(1 + \frac{x}{2} = 1.03\), so \(x = 0.06\). It is small, so the first few terms give a good estimate.
5. Find the coefficient of \(x\) in the expansion of \((1 + x)(1 - x)^7\).
\((1 - x)^7 = 1 - 7x + \dots\). The \(x\) terms come from \(1 \times (-7x)\) and \(x \times 1\), so the coefficient is \(-7 + 1 = -6\).
6. Find the coefficient of \(x^3\) in the expansion of \((2 - x)^7\).
\(\binom73 \times 2^4 \times (-1)^3 = 35 \times 16 \times (-1)\). The coefficient is \(-560\).
Specification checklist
- (P2 4.5) I can work with factorials and the notation \(\binom{n}{r}\), evaluate it with and without a calculator, and explain it as a number of choices.
- (P2 4.5) I can build Pascal's triangle with its addition rule and read off the coefficients of \((a + b)^n\).
- (P2 4.5) I can expand \((a + bx)^n\) for a positive integer \(n\), in full or as its first few terms in ascending powers of \(x\), keeping signs and brackets correct.
- (P2 4.5) I can find one particular term or coefficient using the general term, and give it exactly.
- (P2 4.5) I can set up and solve equations to find an unknown constant, or the power \(n\), from given coefficients, rejecting values that do not fit the conditions.
- (P2 4.5) I can choose a suitable value of \(x\) and use the first few terms of an expansion to approximate a power such as \(0.995^{10}\).
- (P2 4.5) I can find coefficients in the product of a short bracket and a binomial expansion by collecting every pair of terms that gives the required power.