Binomial expansion at A-Level
You already know how to expand brackets like \((x+3)^2\) by multiplying out term by term. Binomial expansion is the same idea scaled up to any power \(n\), using a formula instead of repeated multiplication — essential once expanding by hand would take too long to be practical in an exam.
Four skills make up this topic:
- Pascal's triangle — reading off coefficients directly for small, whole-number powers
- Binomial coefficients — using \(\binom{n}{r}\) (also written \(^nC_r\)) to find coefficients for any power
- Finding a specific term — getting one coefficient without expanding everything
- Binomial estimation — using a truncated expansion to estimate a numerical value
Pascal's triangle
Each row of Pascal's triangle gives the coefficients for expanding \((a+b)^n\), starting at \(n=0\). Every number is the sum of the two above it:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
For small \(n\), this is the fastest way to expand — no formula needed, just read off the row.
Worked Example 1
Use Pascal's triangle to expand \((x+2)^4\).
1
The row of Pascal's triangle for \(n=4\) is \(1, 4, 6, 4, 1\)
2
Pair each coefficient with a term \(x^{4-r}2^r\), decreasing the power of \(x\) and increasing the power of \(2\) each time: \(1(x^4) + 4(x^3 \cdot 2) + 6(x^2\cdot 2^2) + 4(x\cdot 2^3) + 1(2^4)\)
3
Evaluate each term: \(x^4 + 8x^3 + 24x^2 + 32x + 16\)
Answer\((x+2)^4 = x^4+8x^3+24x^2+32x+16\)
Binomial coefficients and the general expansion
For larger \(n\), writing out Pascal's triangle row by row is slow. The binomial coefficient formula gives any entry directly: \(\binom{n}{r} = \dfrac{n!}{r!(n-r)!}\), where \(n! = n \times (n-1) \times \cdots \times 2 \times 1\). The full binomial theorem is:
\[(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r}b^r\]
Worked Example 2
Use the binomial expansion to expand \((2+x)^5\) in ascending powers of \(x\).
1
Find each coefficient: \(\binom{5}{0}=1\), \(\binom{5}{1}=5\), \(\binom{5}{2}=10\), \(\binom{5}{3}=10\), \(\binom{5}{4}=5\), \(\binom{5}{5}=1\)
2
Pair each with \(2^{5-r}x^r\): \(1(2^5) + 5(2^4x) + 10(2^3x^2) + 10(2^2x^3) + 5(2x^4) + 1(x^5)\)
3
Evaluate: \(32 + 80x + 80x^2 + 40x^3 + 10x^4 + x^5\)
Answer\((2+x)^5 = 32+80x+80x^2+40x^3+10x^4+x^5\)
Expanding up to a given term
Exam questions often only need the first few terms, not the full expansion — this saves time and is exactly what you need for approximation later. Work through the same formula, but stop once you reach the term asked for.
Worked Example 3
Find the first four terms, in ascending powers of \(x\), of the expansion of \((1-2x)^6\).
1
Coefficients needed: \(\binom{6}{0}=1\), \(\binom{6}{1}=6\), \(\binom{6}{2}=15\), \(\binom{6}{3}=20\)
2
Pair each with \((-2x)^r\): \(1(1) + 6(-2x) + 15(-2x)^2 + 20(-2x)^3\)
3
Evaluate, taking care with the signs: \(1 - 12x + 60x^2 - 160x^3\)
Answer\(1 - 12x + 60x^2 - 160x^3\)
Finding a specific term
To find one particular coefficient, there's no need to expand everything else — substitute the single value of \(r\) that gives the term you want directly into \(\binom{n}{r}a^{n-r}b^r\).
Worked Example 4
Find the coefficient of \(x^3\) in the expansion of \((3+2x)^7\).
1
The term in \(x^3\) needs \(r=3\): \(\binom{7}{3}(3)^{7-3}(2x)^3\)
2
\(\binom{7}{3}=35\), \(3^4=81\), \(2^3=8\)
3
Coefficient \(= 35 \times 81 \times 8 = 22680\)
AnswerCoefficient of \(x^3\) is \(22680\)
Solving for an unknown coefficient
If a question gives you the value of a coefficient and asks for an unknown inside the bracket, write the general term for that power of \(x\), set it equal to the given value, and solve.
Worked Example 5
The coefficient of \(x^2\) in the expansion of \((1+kx)^5\) is \(90\). Find the possible value(s) of \(k\).
1
The \(x^2\) term is \(\binom{5}{2}(kx)^2 = 10k^2x^2\)
2
Set the coefficient equal to 90: \(10k^2 = 90 \implies k^2 = 9\)
3
Take the square root of both sides — remember both signs
Answer\(k = 3\) or \(k = -3\)
Binomial estimation
A truncated expansion of \((1+x)^n\) with a small value of \(x\) gives a fast, accurate estimate of a numerical power — each extra term contributes far less than the one before, so stopping early barely affects the result.
Worked Example 6
Use the binomial expansion of \((1+x)^5\), up to and including the term in \(x^3\), to estimate \(1.01^5\), giving your answer to 5 decimal places.
1
Write \(1.01 = 1+x\) with \(x=0.01\)
2
Expand up to \(x^3\): \((1+x)^5 \approx 1+5x+10x^2+10x^3\)
3
Substitute \(x=0.01\): \(1 + 0.05 + 0.001 + 0.00001\)
Answer\(1.01^5 \approx 1.05101\)
Exam tips
💡 Tip 1
Write the general term before substituting anything
\(\binom{n}{r}a^{n-r}b^r\) as a labelled line of working, with your values of \(n\), \(a\), \(b\) and \(r\) written in, catches substitution errors before they turn into a wrong final answer.
💡 Tip 2
Watch the sign when \(b\) is negative
In \((1-2x)^n\), the "\(b\)" term is \(-2x\), not \(2x\) — every odd power of \(r\) then carries a negative sign. Substitute \((-2x)^r\) as a whole, not \(-(2x)^r\), to avoid losing track of it.
💡 Tip 3
For "find the coefficient", give a number — not a term
If asked for "the coefficient of \(x^3\)", the answer is a number (e.g. \(22680\)), not \(22680x^3\) — including the \(x^3\) in the final answer line is a common way to lose an easy mark.
💡 Tip 4
Square-root equations for \(k\) almost always have two solutions
Solving \(k^2 = 9\) gives \(k = \pm 3\) — dropping the negative root is one of the most common lost marks in this topic, unless the question states a restriction (e.g. "\(k>0\)").
💡 Tip 5
For estimation, match your expansion to the value of \(x\)
The technique only works well when \(x\) is small (e.g. \(0.01\) or \(0.02\)) — this is why \(1.01^5\) is written as \((1+x)^5\) with \(x=0.01\), not restructured some other way.
Common mistakes
Common Mistake 1
Forgetting to raise the whole bracketed term to the power \(r\)
In \(\binom{5}{3}(2x)^3\), both the 2 and the \(x\) must be cubed: \((2x)^3 = 8x^3\), not \(2x^3\). Treating \(2x\) as if only \(x\) were raised to the power is a very common slip.
Common Mistake 2
Using \(\binom{n}{r}\) with the wrong value of \(r\)
The power of \(x\) in a term always equals \(r\), not \(n-r\) — for the \(x^3\) term, use \(r=3\) throughout, in both the binomial coefficient and the powers of \(a\) and \(b\).
Common Mistake 3
Mixing up Pascal's triangle rows
The row for \((a+b)^n\) is the \((n+1)\)th row of the triangle, since the first row (a single 1) corresponds to \(n=0\) — an easy off-by-one error when counting rows by hand.
Common Mistake 4
Expanding fully when only one term was asked for
Finding a single specific term doesn't require the rest of the expansion — writing out every term to find just one wastes exam time and adds unnecessary chances for arithmetic slips.
Common Mistake 5
Rounding too early in an estimation question
Round only the final answer, to the precision asked for — rounding each term of the expansion first before adding them introduces small errors that can shift the final decimal place.
Practice questions
Work through each question before checking the worked solution.
Core Skills
Q1Use Pascal's triangle to expand \((x+1)^3\).Core Skills
Q2Evaluate \(\binom{6}{2}\).Core Skills
Q3Write down the row of Pascal's triangle for \(n=5\).Core Skills
Q4Use the binomial expansion to expand \((x+3)^4\) fully.Core Skills
Q5Find the value of \(\binom{8}{3}\).Core Skills
Exam-Style
Q6Find the first three terms, in ascending powers of \(x\), of the expansion of \((1+3x)^6\).Exam-Style
Q7Find the coefficient of \(x^2\) in the expansion of \((2-x)^6\).Exam-Style
Q8Expand \((2x-1)^5\) fully.Exam-Style
Q9Given that the coefficient of \(x\) in the expansion of \((1+ax)^4\) is \(12\), find the value of \(a\).Exam-Style
Q10Use the binomial expansion of \((1+x)^4\), up to and including the term in \(x^2\), to estimate \(1.02^4\) correct to 4 decimal places.Exam-Style
A* Challenge
Q11Find the term independent of \(x\) in the expansion of \(\left(2x+\dfrac{1}{x}\right)^6\).A* Challenge
Q12Find the coefficient of \(x^3\) in the expansion of \((1+x)(1+2x)^5\).A* Challenge
Q13The first three terms, in ascending powers of \(x\), of the expansion of \((1+px)^n\) are \(1-14x+84x^2-\ldots\). Find the values of \(n\) and \(p\).A* Challenge
Q14Find the coefficient of \(x^4\) in the expansion of \((1+2x)^3(1-x)^5\).A* Challenge
Q15In the expansion of \((1+x)^n\), the coefficient of \(x^3\) is twice the coefficient of \(x^2\). Find the value of \(n\).A* Challenge
Answers — full worked solutions
Core Skills (Q1–Q5)
Q1 — \(x^3+3x^2+3x+1\)
1
Row for \(n=3\): \(1, 3, 3, 1\)
\((x+1)^3 = x^3+3x^2+3x+1\)
Q2 — \(15\)
1
\(\binom{6}{2} = \dfrac{6!}{2!4!} = \dfrac{6\times5}{2\times1}\)
\(\binom{6}{2} = 15\)
Q3 — \(1, 5, 10, 10, 5, 1\)
1
Add adjacent pairs from the \(n=4\) row \(1,4,6,4,1\)
\(1, 5, 10, 10, 5, 1\)
Q4 — \(x^4+12x^3+54x^2+108x+81\)
1
Row for \(n=4\): \(1,4,6,4,1\), paired with \(x^{4-r}3^r\)
2
\(x^4 + 4(3)x^3 + 6(9)x^2 + 4(27)x + 81\)
\(x^4+12x^3+54x^2+108x+81\)
Q5 — \(56\)
1
\(\binom{8}{3} = \dfrac{8!}{3!5!} = \dfrac{8\times7\times6}{3\times2\times1}\)
\(\binom{8}{3} = 56\)
Exam-Style (Q6–Q10)
Q6 — \(1+18x+135x^2\)
1
\(\binom{6}{0}=1\), \(\binom{6}{1}(3)=18\), \(\binom{6}{2}(3)^2=15\times9=135\)
\(1+18x+135x^2\)
Q7 — \(240\)
1
\(x^2\) term: \(\binom{6}{2}(2)^4(-1)^2 = 15\times16\times1\)
Coefficient \(= 240\)
Q8 — \(32x^5-80x^4+80x^3-40x^2+10x-1\)
1
Terms: \(\binom{5}{r}(2x)^{5-r}(-1)^r\) for \(r=0\) to \(5\)
2
\(32x^5,\ -80x^4,\ 80x^3,\ -40x^2,\ 10x,\ -1\)
\(32x^5-80x^4+80x^3-40x^2+10x-1\)
Q9 — \(a=3\)
1
\(x\) term: \(\binom{4}{1}a = 4a\)
2
\(4a=12\)
\(a=3\)
Q10 — \(1.0824\)
1
\((1+x)^4 \approx 1+4x+6x^2\), with \(x=0.02\)
2
\(1+0.08+6(0.0004) = 1+0.08+0.0024\)
\(1.02^4 \approx 1.0824\)
A* Challenge (Q11–Q15)
Q11 — \(160\)
1
General term: \(\binom{6}{r}(2x)^{6-r}(x^{-1})^r = \binom{6}{r}2^{6-r}x^{6-2r}\)
2
Independent of \(x\) means \(6-2r=0 \implies r=3\)
3
\(\binom{6}{3}2^3 = 20\times8\)
Constant term \(=160\)
Q12 — \(120\)
1
Coefficient of \(x^3\) in \((1+2x)^5\): \(\binom{5}{3}2^3 = 10\times8=80\)
2
Coefficient of \(x^2\) in \((1+2x)^5\): \(\binom{5}{2}2^2 = 10\times4=40\)
3
\((1+x)(1+2x)^5\) contributes the \(x^2\) coefficient (shifted up by the extra \(x\)) plus the \(x^3\) coefficient: \(80+40\)
Coefficient of \(x^3\) is \(120\)
Q13 — \(n=7\), \(p=-2\)
1
\(x\) coefficient: \(np=-14\)
2
\(x^2\) coefficient: \(\binom{n}{2}p^2=84 \implies n(n-1)p^2=168\)
3
Substitute \(p=-\dfrac{14}{n}\): \(196(n-1) = 168n \implies 28n=196\)
\(n=7\), \(p=-2\)
Q14 — \(25\)
1
\((1+2x)^3 = 1+6x+12x^2+8x^3\); \((1-x)^5 = 1-5x+10x^2-10x^3+5x^4-x^5\)
2
Coefficient of \(x^4\) in the product: \((1)(5)+(6)(-10)+(12)(10)+(8)(-5)\)
3
\(5-60+120-40\)
Coefficient of \(x^4\) is \(25\)
Q15 — \(n=8\)
1
\(\binom{n}{3} = 2\binom{n}{2} \implies \dfrac{n(n-1)(n-2)}{6} = 2\times\dfrac{n(n-1)}{2}\)
2
Divide both sides by \(n(n-1)\): \(\dfrac{n-2}{6} = 1\)
\(n=8\)
Summary
- Pascal's triangle gives coefficients directly for small \(n\); for larger \(n\), use \(\binom{n}{r} = \dfrac{n!}{r!(n-r)!}\).
- The full expansion is \((a+b)^n = \sum_{r=0}^{n}\binom{n}{r}a^{n-r}b^r\) — always raise the whole bracketed term \(b\) to the power \(r\), not just the variable inside it.
- To find one specific term, substitute the single matching value of \(r\) — there's no need to expand the rest.
- To solve for an unknown coefficient, write the general term for that power of \(x\) and set it equal to the given value.
- Binomial estimation truncates the expansion of \((1+x)^n\) for small \(x\) — round only the final answer, not each term along the way.
This topic builds on the expanding-brackets skills from GCSE Expanding Brackets and GCSE Quadratic Expressions, and the algebraic rigour from Algebraic Methods. Next in Pure Year 1: Trigonometric Ratios.
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