What is the Binomial Theorem and how is it used in Edexcel International A Levels Pure Mathematics 2?

The Binomial Theorem is a fundamental principle in algebra that provides a formula for expanding expressions raised to a power, specifically of the form (a + b)^n. In Edexcel International A Levels Pure Mathematics 2, it is used to expand binomials into a series, which is particularly useful for simplifying complex algebraic expressions and solving problems involving powers and coefficients.

How do you apply the Binomial Theorem to expand (x + y)^n in Pure Mathematics 2?

To expand (x + y)^n using the Binomial Theorem, you use the formula: (x + y)^n = Σ [n choose k] * x^(n-k) * y^k, where 'Σ' denotes the sum, 'n choose k' is the binomial coefficient calculated as n!/(k!(n-k)!), and k ranges from 0 to n. This expansion is a key topic in Edexcel International A Levels Pure Mathematics 2, allowing students to express the binomial as a polynomial.

What are binomial coefficients and how do you calculate them in the context of the Binomial expansion?

Binomial coefficients are the numerical factors that multiply the terms in the expansion of a binomial raised to a power, represented as 'n choose k' or C(n, k). They are calculated using the formula n!/(k!(n-k)!), where '!' denotes factorial. Understanding how to compute these coefficients is crucial for mastering Binomial expansion in Edexcel International A Levels Pure Mathematics 2.

Can you explain the concept of a binomial expansion's general term in Pure Mathematics 2?

The general term in a binomial expansion, often referred to as the 'k-th term', is given by T(k+1) = [n choose k] * x^(n-k) * y^k. This formula allows students to find any specific term in the expansion without fully expanding the binomial. This concept is particularly useful in Edexcel International A Levels Pure Mathematics 2 for efficiently solving problems involving large powers.

How does understanding Binomial expansion help in solving real-world problems in Mathematics?

Understanding Binomial expansion is essential for solving real-world problems that involve probability, statistics, and algebraic simplifications. It allows students to expand expressions efficiently, calculate probabilities in binomial distributions, and analyze polynomial functions. Mastery of this topic in Edexcel International A Levels Pure Mathematics 2 equips students with the skills to tackle complex mathematical challenges in various fields.

Why is "Forgetting to raise the coefficient inside the bracket to the power" a common mistake in Binomial expansion?

Why it happens: In (2 + 3x)^4 students often compute 42 · 2 · (3x)^2 = 6 · 2 · 9x^2 — instead of 42 · 2^2 · (3x)^2. How to avoid it: In nr a^n - r b^r both a AND b must be raised to their respective powers, INCLUDING the numerical part of b. So (3x)^2 = 9 x^2, and the term is 6 · 4 · 9 x^2 = 216 x^2. Source: WMA12/01 examiner reports — flagged annually

Why is "Losing the minus sign when expanding $(a - b)^{n}$" a common mistake in Binomial expansion?

Why it happens: Students treat (a - b)^n as (a + b)^n and forget the alternating signs. How to avoid it: Write (a - b)^n as (a + (-b))^n. Odd-power terms in b get a minus sign; even-power terms keep a plus. E.g. (2 - x)^4 = 16 - 32x + 24 x^2 - 8 x^3 + x^4.

Why is "Setting $r = 4$ to find the term in $x^{4}$ in $(2 + x)^{6}$ — and getting the wrong term" a common mistake in Binomial expansion?

Why it happens: Confusion between term index (T_r + 1) and exponent. How to avoid it: The exponent of b equals r. So in (a + b)^n the term containing b^4 has r = 4, and it's the 5-th term (since the first term has r = 0). When unsure, write the general term out and set the b-power to the value you want.

Why is "Applying $\binom{n}{r}$ when $n = -2$ or $n = 1/3$" a common mistake in Binomial expansion?

Why it happens: Confusing P2's binomial expansion (positive integer n) with P4's generalisation. How to avoid it: In P2 (WMA12) the binomial theorem is ONLY for positive integer n. The expansion is finite. Negative and fractional exponents (with their infinite series and convergence conditions) are studied in P4 (WMA14).

Why is "Writing the expansion in descending powers when the question says 'ascending'" a common mistake in Binomial expansion?

Why it happens: Habit from algebra (writing polynomials in standard form). How to avoid it: If the question says 'ascending powers of x', start with the constant and increase the exponent. Examiners may not award marks if the order is wrong, especially in 'first three terms' style questions.

Edexcel International A Levels Mathematics (XMA01-YMA01)

Binomial expansion

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Short Notes - Binomial expansion

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Detailed Study Notes

Detailed notes on Pure Mathematics 2 for Edexcel International A Levels Mathematics, covering key concepts, explanations, examples, and exam-focused revision points.

Binomial Expansion Study Notes — Edexcel IAL Mathematics P2 (WMA12/01, 2018 spec — 2026 onwards)

Expansion of $(a + b)^{n}$ for positive integer $n$ . Pascal's triangle, the binomial coefficient $\binom{n}{r}$ , the general term, and approximation problems. The finite-series case — fractional and negative exponents are studied in P4.

At a glance

Binomial theorem: $(a + b)^{n} = \sum_{r = 0}^{n} \binom{n}{r} a^{n - r} b^{r}$ .
Binomial coefficient: $\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$ .
General term: $T_{r + 1} = \binom{n}{r} a^{n - r} b^{r}$ (note the $r + 1$ — first term has $r = 0$ ).
Pascal's triangle: each row $n$ gives the coefficients $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ .
Special case: $(1 + x)^{n} = 1 + nx + \binom{n}{2} x^{2} + \ldots + x^{n}$ .
P2 restriction: $n$ is a positive integer. Negative/fractional exponents are P4 territory.
Formula booklet: $\binom{n}{r}$ and $(a + b)^{n}$ ARE given.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XMA01-YMA01 syllabus (2026 onwards).

P2 4.1 — Understand and use the binomial expansion of $(a + b)^{n}$ for positive integer $n$ .
P2 4.2 — Use the notations $n!$ and $\binom{n}{r}$ (or ${}^{n}C_{r}$ ).
P2 4.3 — Use Pascal's triangle for small values of $n$ .
P2 4.4 — Use the binomial expansion to find approximate values, e.g. $1.05^{10}$ .

Pascal's triangle and binomial coefficients

Each row is built by summing pairs from the row above.

Pascal's triangle.

$n$	Row
$0$	$1$
$1$	$1 \; 1$
$2$	$1 \; 2 \; 1$
$3$	$1 \; 3 \; 3 \; 1$
$4$	$1 \; 4 \; 6 \; 4 \; 1$
$5$	$1 \; 5 \; 10 \; 10 \; 5 \; 1$
$6$	$1 \; 6 \; 15 \; 20 \; 15 \; 6 \; 1$

Each interior entry is the sum of the two entries above. Row $n$ gives the binomial coefficients $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ .

Pascal's triangle: row n holds the binomial coefficients C(n, r) for (a + b)ⁿ, each interior entry equalling the pair above.

Binomial coefficient formula.

$\binom{n}{r} = \dfrac{n!}{r!(n - r)!} = \dfrac{n(n - 1)(n - 2) \cdots (n - r + 1)}{r!}.$

Worked computation. $\binom{8}{3} = \dfrac{8 \cdot 7 \cdot 6}{3!} = \dfrac{336}{6} = 56$ .

Properties.

$\binom{n}{0} = \binom{n}{n} = 1$ .
$\binom{n}{1} = \binom{n}{n - 1} = n$ .
Symmetry: $\binom{n}{r} = \binom{n}{n - r}$ .
Sum of row: $\sum_{r = 0}^{n} \binom{n}{r} = 2^{n}$ (set $x = 1$ in $(1 + x)^{n}$ ).

Row $n$ has $n + 1$ entries.
$\binom{n}{r} = \binom{n}{n - r}$ (symmetry).
$\binom{n}{0} = 1$ , $\binom{n}{1} = n$ , $\binom{n}{n} = 1$ .
Sum of row $n$ is $2^{n}$ .

The binomial theorem

Expand $(a + b)^{n}$ term by term using $\binom{n}{r} a^{n - r} b^{r}$ .

Statement.

$(a + b)^{n} = \sum_{r = 0}^{n} \binom{n}{r} a^{n - r} b^{r} = a^{n} + \binom{n}{1} a^{n - 1} b + \binom{n}{2} a^{n - 2} b^{2} + \ldots + b^{n}.$

Each term is $\binom{n}{r} a^{n - r} b^{r}$ — the power of $a$ decreases from $n$ to $0$ and the power of $b$ increases from $0$ to $n$ as $r$ goes from $0$ to $n$ .

Worked example. Expand $(2 + 3x)^{4}$ fully.

Row 4 of Pascal: $1, 4, 6, 4, 1$ .

$(2 + 3x)^{4} = 1 \cdot 2^{4} + 4 \cdot 2^{3} (3x) + 6 \cdot 2^{2} (3x)^{2} + 4 \cdot 2 (3x)^{3} + 1 \cdot (3x)^{4}$

$= 16 + 96x + 216 x^{2} + 216 x^{3} + 81 x^{4}.$

Critical: $(3x)^{2} = 9 x^{2}$ , not $3 x^{2}$ . The whole bracket gets raised to the power.

Special case $(1 + x)^{n}$ . Set $a = 1$ , $b = x$ :

$(1 + x)^{n} = 1 + n x + \dfrac{n(n - 1)}{2!} x^{2} + \dfrac{n(n - 1)(n - 2)}{3!} x^{3} + \ldots + x^{n}.$

This form is THE most common in Edexcel questions — many problems are set up so the bracket is $(1 + \text{something})^{n}$ .

Ascending vs descending powers. Edexcel exams almost always ask for terms in ascending powers of $x$ — start with the constant, then $x$ , then $x^{2}$ , etc. Read the question carefully.

Each term: $\binom{n}{r} a^{n - r} b^{r}$ .
Powers of $a$ go DOWN, powers of $b$ go UP.
Always raise the WHOLE bracket to the power.
Watch out: 'ascending' means constant first.

General term — finding a specific coefficient

$T_{r + 1} = \binom{n}{r} a^{n - r} b^{r}$ . Match the power of $b$ to the term you want.

General term formula.

$T_{r + 1} = \binom{n}{r} a^{n - r} b^{r}.$

(Note: the $(r + 1)$ -th term has the $b^{r}$ — because $T_{1}$ has $r = 0$ .)

Strategy. To find the coefficient of $x^{k}$ in $(a + b)^{n}$ where $b = cx$ for some constant $c$ :

Identify which $r$ gives the power $x^{k}$ . Usually $r = k$ .
Substitute into the general term and compute.

Worked example. Find the coefficient of $x^{5}$ in $(3 - 2x)^{8}$ .

$a = 3$ , $b = -2x$ , $n = 8$ . The $x^{5}$ term has $r = 5$ .
$T_{6} = \binom{8}{5} \cdot 3^{3} \cdot (-2x)^{5} = 56 \cdot 27 \cdot (-32) x^{5} = -48384 x^{5}$ .
Coefficient: $-48384$ .

Sign warning. When $b$ has a minus sign, the parity of $r$ determines the overall sign. Odd $r$ → negative; even $r$ → positive.

Cross-coefficient ratios. A standard exam trick:

$\dfrac{T_{r + 2}}{T_{r + 1}} = \dfrac{\binom{n}{r + 1}}{\binom{n}{r}} \cdot \dfrac{b}{a} = \dfrac{n - r}{r + 1} \cdot \dfrac{b}{a}.$

This lets you set up equations to find $n$ or $k$ when two coefficients are given.

Worked example. In $(1 + kx)^{n}$ , the coefficient of $x^{2}$ is $90$ and of $x^{3}$ is $270$ . Find $k$ , $n$ .

Ratio: $\dfrac{\binom{n}{3} k^{3}}{\binom{n}{2} k^{2}} = \dfrac{n - 2}{3} k = \dfrac{270}{90} = 3$ , so $(n - 2) k = 9$ .
Substituting: $\binom{n}{2} k^{2} = 90$ becomes $\dfrac{n(n - 1)}{2} \cdot \dfrac{81}{(n - 2)^{2}} = 90$ .
Solve: $n = 5$ , $k = 3$ .

$T_{r + 1}$ contains $b^{r}$ — off by one!
Match the power of $b$ to the desired term.
Sign of $T_{r + 1}$ depends on the parity of $r$ when $b < 0$ .
Ratio trick: $T_{r + 2}/T_{r + 1}$ simplifies algebra hugely.

Approximation using binomial expansion

Use the first few terms with a small $x$ to approximate $(1 + x)^{n}$ .

Idea. When $|x|$ is small, the first few terms of $(1 + x)^{n}$ give an excellent approximation. The later terms (in $x^{k}$ for large $k$ ) contribute negligibly.

Procedure.

Write the expression you want to approximate in the form $(1 + \text{small})^{n}$ .
Expand to the requested number of terms.
Substitute the small value for $x$ and compute.

Worked example. Approximate $0.98^{5}$ using the first three terms of $(1 - 2x)^{5}$ .

$0.98 = 1 - 0.02 = 1 - 2(0.01)$ , so set $x = 0.01$ .
$(1 - 2x)^{5} = 1 + 5(-2x) + \binom{5}{2}(-2x)^{2} + \ldots = 1 - 10x + 40 x^{2} + \ldots$
Plug in: $1 - 0.1 + 0.004 = 0.904$ .
Exact value: $0.98^{5} \approx 0.903921$ . Good to $4$ s.f.

How many terms to include? Usually the question tells you ('first three terms', 'up to and including $x^{2}$ '). If not, judge by the required accuracy: each additional term improves the approximation by roughly a factor of $x$ .

Choosing the substitution. Most numerical-approximation questions have the substitution chosen so that $x$ is a 'nice' number like $0.01$ or $0.1$ . If the substitution gives $x = 0.5$ , the approximation will be poor — the question is likely asking for the wrong form.

Identify the form $(1 + \text{small})^{n}$ .
Expand to the requested number of terms.
Substitute the small value.
Approximation accuracy improves with smaller $x$ and more terms.

Quick recap

$(a + b)^{n} = \sum_{r = 0}^{n} \binom{n}{r} a^{n - r} b^{r}$ for positive integer $n$ .
$\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$ ; symmetric: $\binom{n}{r} = \binom{n}{n - r}$ .
Pascal's triangle: row $n$ gives the coefficients.
General term: $T_{r + 1}$ contains $b^{r}$ .
$(1 + x)^{n} = 1 + nx + \binom{n}{2} x^{2} + \ldots + x^{n}$ .
Sign of terms in $(a - b)^{n}$ alternates.
Approximation: small $x$ + first few terms.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Binomial theorem: $(a + b)^{n} = \sum \binom{n}{r} a^{n - r} b^{r}$ .
$\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$ .
$\binom{n}{0} = \binom{n}{n} = 1$ , $\binom{n}{1} = n$ .
Pascal's rule: each interior entry is the sum of the two above.
$(1 + x)^{n}$ shortcut: $1 + nx + \dfrac{n(n - 1)}{2} x^{2} + \ldots$

How it’s examined

Binomial expansion appears in WMA12/01 P2 every sitting (Jan & Jun), typically as Q2-Q11 — worth $5$ - $10$ marks. Typical questions: expand a binomial fully (4 marks); find the first three or four terms in ascending powers (5 marks); find the coefficient of a specific term using $T_{r + 1}$ (4-5 marks); find unknowns $n$ and $k$ from two given coefficients (6-7 marks); use a binomial expansion to approximate a numerical value (4-5 marks). Mark schemes apply ECF liberally between consecutive terms. A* students should aim for full marks — these are highly mechanical once the structure is internalised.

Sources: Pearson Edexcel International Advanced Level Mathematics specification (2018 — current for 2026 onwards); Edexcel IAL Mathematical Formulae and Statistical Tables (2018); WMA12/01 Examiner Reports — January 2023, June 2023, January 2024, June 2024. Last reviewed 2026-05-12.

Take this whole topic with you

Step-by-step worked examples — Binomial expansion

Step-by-step solutions to past-paper-style questions on binomial expansion, written exactly the way a tutor would explain them at the board.

1Expand a binomial fully (4 marks, P2)
Core• Adapted from WMA12/01 January 2024 Q2• binomial expansion, Pascal's triangle
Question
Use the binomial theorem to expand $(2 + 3x)^{4}$ in ascending powers of $x$ up to and including the term in $x^{4}$ .
Step-by-step solution
1. Step 1
  Use $(a + b)^{n} = \sum_{r=0}^{n} \binom{n}{r} a^{n - r} b^{r}$ with $a = 2$ , $b = 3x$ , $n = 4$ .
2. Step 2
  Coefficients from Pascal's triangle (row $n = 4$ ): $1, 4, 6, 4, 1$ .
3. Step 3
  Term-by-term:
  $1 \cdot 2^{4} + 4 \cdot 2^{3}(3x) + 6 \cdot 2^{2}(3x)^{2} + 4 \cdot 2(3x)^{3} + 1 \cdot (3x)^{4}$
4. Step 4
  Compute each: $16 + 4 \cdot 8 \cdot 3x + 6 \cdot 4 \cdot 9 x^{2} + 4 \cdot 2 \cdot 27 x^{3} + 81 x^{4}$ .
5. Step 5
  Simplify:
  $16 + 96x + 216 x^{2} + 216 x^{3} + 81 x^{4}$
Answer
$16 + 96x + 216 x^{2} + 216 x^{3} + 81 x^{4}$ .
Examiner tip
M1 for using the binomial structure with all five terms attempted. A1 for the constant ( $16$ ). A1 for the $x$ coefficient ( $96$ ). A1 for the $x^{2}$ coefficient ( $216$ ). The remaining two terms typically share an A mark in shorter questions. Common slip: forgetting that the cube of $3x$ is $27 x^{3}$ (students write $9 x^{3}$ ).
2First four terms of $(1 + kx)^{n}$ (5 marks, P2)
Core• Adapted from WMA12/01 June 2024 Q3• binomial expansion, first four terms
Question
Find the first four terms, in ascending powers of $x$ , of the binomial expansion of $(1 + 2x)^{10}$ .
Step-by-step solution
1. Step 1
  Apply $(1 + y)^{n} = 1 + \binom{n}{1} y + \binom{n}{2} y^{2} + \binom{n}{3} y^{3} + \ldots$ with $y = 2x$ , $n = 10$ .
2. Step 2
  Compute the coefficients: $\binom{10}{1} = 10$ , $\binom{10}{2} = 45$ , $\binom{10}{3} = 120$ .
3. Step 3
  Term in $x$ : $10 \cdot 2x = 20x$ . Term in $x^{2}$ : $45 \cdot (2x)^{2} = 45 \cdot 4 x^{2} = 180 x^{2}$ . Term in $x^{3}$ : $120 \cdot (2x)^{3} = 120 \cdot 8 x^{3} = 960 x^{3}$ .
4. Step 4
  Assemble:
  $1 + 20x + 180 x^{2} + 960 x^{3} + \ldots$
Answer
$1 + 20x + 180 x^{2} + 960 x^{3}$ (in ascending powers of $x$ ).
Examiner tip
B1 for the constant term ( $1$ ). M1 for the binomial structure with $\binom{n}{r}$ used correctly. A1 for $20x$ . A1 for $180 x^{2}$ . A1 for $960 x^{3}$ . ECF applies — a slip in one term doesn't penalise subsequent terms.
3Find the coefficient of a particular term (5 marks, P2)
Extended• Adapted from WMA12/01 January 2023 Q5• binomial coefficient, general term
Question
Find the coefficient of $x^{5}$ in the binomial expansion of $(3 - 2x)^{8}$ .
Step-by-step solution
1. Step 1
  General term: $T_{r+1} = \binom{8}{r} (3)^{8 - r} (-2x)^{r}$ .
2. Step 2
  Want the $x^{5}$ term, so $r = 5$ :
  $T_{6} = \binom{8}{5} \cdot 3^{3} \cdot (-2)^{5} \cdot x^{5}$
3. Step 3
  Compute: $\binom{8}{5} = 56$ ; $3^{3} = 27$ ; $(-2)^{5} = -32$ .
4. Step 4
  Multiply the constants: $56 \cdot 27 \cdot (-32) = 1512 \cdot (-32) = -48384$ .
Answer
Coefficient of $x^{5}$ is $-48384$ .
Examiner tip
M1 for the general-term structure with $r = 5$ . A1 for $\binom{8}{5} = 56$ . A1 for $3^{3} \cdot (-2)^{5} = -864$ . A1 for $-48384$ . The minus sign is a frequent mark-loser — $(-2)^{5}$ is odd-power, so negative.
4Find $n$ from a given coefficient (6 marks, P2)
Extended• Adapted from WMA12/01 June 2023 Q6• finding n, coefficients
Question
In the expansion of $(1 + kx)^{n}$ , where $k$ is a positive constant and $n$ is a positive integer, the coefficient of $x^{2}$ is $90$ and the coefficient of $x^{3}$ is $270$ . Find the values of $k$ and $n$ .
Step-by-step solution
1. Step 1
  Coefficient of $x^{2}$ : $\binom{n}{2} k^{2} = 90$ , i.e. $\dfrac{n(n - 1)}{2} k^{2} = 90$ . Coefficient of $x^{3}$ : $\binom{n}{3} k^{3} = 270$ , i.e. $\dfrac{n(n - 1)(n - 2)}{6} k^{3} = 270$ .
2. Step 2
  Take the ratio (third coefficient over second):
  $\dfrac{\binom{n}{3} k^{3}}{\binom{n}{2} k^{2}} = \dfrac{n - 2}{3} \cdot k = \dfrac{270}{90} = 3$
3. Step 3
  So $(n - 2) k = 9$ , i.e. $k = \dfrac{9}{n - 2}$ . ( $\star$ )
4. Step 4
  Substitute back into $\binom{n}{2} k^{2} = 90$ :
  $\dfrac{n(n - 1)}{2} \cdot \dfrac{81}{(n - 2)^{2}} = 90$
5. Step 5
  Rearrange: $\dfrac{81 n(n - 1)}{2 (n - 2)^{2}} = 90$ , so $81 n(n - 1) = 180 (n - 2)^{2}$ .
6. Step 6
  Expand: $81 n^{2} - 81 n = 180 (n^{2} - 4n + 4) = 180 n^{2} - 720 n + 720$ . Bring all to one side: $99 n^{2} - 639 n + 720 = 0$ . Divide by $3$ : $33 n^{2} - 213 n + 240 = 0$ , divide by $3$ again: $11 n^{2} - 71 n + 80 = 0$ .
7. Step 7
  Use the quadratic formula: $n = \dfrac{71 \pm \sqrt{71^{2} - 4 \cdot 11 \cdot 80}}{22} = \dfrac{71 \pm \sqrt{5041 - 3520}}{22} = \dfrac{71 \pm \sqrt{1521}}{22} = \dfrac{71 \pm 39}{22}$ . So $n = 5$ or $n = \tfrac{32}{22} = \tfrac{16}{11}$ .
8. Step 8
  $n$ is a positive integer, so $n = 5$ . Then from ( $\star$ ): $k = \dfrac{9}{5 - 2} = 3$ .
Answer
$n = 5$ , $k = 3$ .
Examiner tip
M1 for both coefficient equations. M1 for the ratio. A1 for $(n-2)k = 9$ . M1 for substituting back. M1 for the resulting quadratic. A1 for both values. Note: only the integer root for $n$ is admissible — the question states $n$ is a positive integer. The ratio trick (taking $C_{r+1} / C_{r}$ ) reduces the algebra dramatically and is a standard Edexcel strategy.
5Approximation using a binomial (4 marks, P2)
Extended• Adapted from WMA12/01 January 2024 Q11• approximation, small x
Question
(a) Find the first three terms, in ascending powers of $x$ , of the expansion of $(1 - 2x)^{5}$ . (b) Hence find an approximation for $0.98^{5}$ , giving your answer to $5$ decimal places.
Step-by-step solution
1. Step 1
  (a) Apply the binomial: $(1 - 2x)^{5} = 1 + 5(-2x) + \binom{5}{2}(-2x)^{2} + \ldots$
2. Step 2
  Simplify the first three terms: $1 - 10x + 10 \cdot 4x^{2} + \ldots = 1 - 10x + 40 x^{2} + \ldots$
3. Step 3
  (b) $0.98 = 1 - 0.02 = 1 - 2(0.01)$ , so set $x = 0.01$ .
4. Step 4
  Approximation: $1 - 10(0.01) + 40(0.01)^{2} = 1 - 0.1 + 0.004 = 0.904$ .
Answer
(a) $1 - 10x + 40 x^{2}$ . (b) $0.98^{5} \approx 0.90400$ .
Examiner tip
(a) M1 for the binomial structure. A1 for $-10x$ . A1 for $+40 x^{2}$ . (b) M1 for substituting $x = 0.01$ . A1 for $0.90400$ . The exact value is $\approx 0.903921$ , so the three-term approximation is good to $4$ s.f. Common slip: students set $x = 0.02$ (rather than $0.01$ ), giving the wrong answer.

Key Formulae — Binomial expansion

The formulae you need to memorise for binomial expansion on the Pearson Edexcel IAL XMA01 / YMA01 paper, with every variable defined in plain English and a note on when to use it.

Binomial theorem (positive integer $n$ )
$(a + b)^{n} = \sum_{r = 0}^{n} \binom{n}{r} a^{n - r} b^{r}$
$a, b$
any real terms
$n$
positive integer
$r$
term index, $0 \le r \le n$
When to use
Expanding $(a + b)^{n}$ in full or finding a specific coefficient. GIVEN in the IAL formula booklet (Pure Mathematics).
Example
$(2 + 3x)^{4} = 16 + 96x + 216 x^{2} + 216 x^{3} + 81 x^{4}$ .
Binomial coefficient
$\binom{n}{r} = \dfrac{n!}{r!(n - r)!}$
$n!$
$n$ factorial, $n \cdot (n - 1) \cdots 2 \cdot 1$
$n$
positive integer
$r$
integer with $0 \le r \le n$
When to use
Computing the coefficient of $a^{n - r} b^{r}$ in the binomial expansion. GIVEN in the IAL formula booklet.
Example
$\binom{10}{3} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120$ .
Special case $(1 + x)^{n}$
$(1 + x)^{n} = 1 + n x + \dfrac{n(n - 1)}{2!} x^{2} + \dfrac{n(n - 1)(n - 2)}{3!} x^{3} + \ldots + x^{n}$
$x$
any real
$n$
positive integer
When to use
The most common exam form. Numerical approximation questions usually arrange the expression so it has the form $(1 + \text{something})^{n}$ . NOT in the formula booklet (but derivable from the general theorem).
Example
$(1 + 2x)^{10} = 1 + 20x + 180 x^{2} + 960 x^{3} + \ldots$
General term ( $T_{r + 1}$ )
$T_{r + 1} = \binom{n}{r} a^{n - r} b^{r}$
$T_{r + 1}$
term containing $b^{r}$
$n, r$
as in binomial theorem
When to use
When asked for a specific term or coefficient without expanding everything. NOT in the formula booklet, but follows directly from the binomial theorem.
Example
In $(3 - 2x)^{8}$ , the $x^{5}$ term has $r = 5$ : $\binom{8}{5} \cdot 3^{3} \cdot (-2)^{5} \cdot x^{5} = -48384 x^{5}$ .

Key Definitions and Keywords — Binomial expansion

Definitions to memorise and the exact keywords mark schemes credit for binomial expansion answers — sharpened from recent examiner reports for the 2026 Pearson Edexcel IAL XMA01 / YMA01 sitting.

Binomial
A two-term algebraic expression of the form $a + b$ . The binomial theorem expands a positive-integer power of a binomial as a finite sum.
Binomial coefficient ${}^{n}C_{r}$ or $\binom{n}{r}$
Examiner keyword
The number of ways of choosing $r$ items from $n$ , equal to $\dfrac{n!}{r!(n - r)!}$ . Also the coefficient of $a^{n - r} b^{r}$ in $(a + b)^{n}$ .
Example
$\binom{8}{3} = \dfrac{8!}{3! 5!} = 56$ .
Pascal's triangle
A triangular array where each entry equals the sum of the two entries above it. Row $n$ gives the binomial coefficients $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ .
Example
Row 4: $1, 4, 6, 4, 1$ — the coefficients in $(a + b)^{4}$ .
Factorial
$n! = n \cdot (n - 1) \cdot (n - 2) \cdots 2 \cdot 1$ for positive integer $n$ , with $0! = 1$ by convention. Calculators have an $n!$ button.
Example
$5! = 120$ .
Ascending powers (of $x$ )
Examiner keyword
Listing terms from constant first, then $x^{1}$ , then $x^{2}$ , etc. — i.e. by increasing exponent. Required wording in WMA12 binomial questions.

Common Mistakes and Misconceptions — Binomial expansion

The traps other students keep falling into on binomial expansion questions — taken from recent Pearson Edexcel IAL XMA01 / YMA01 examiner reports and mark schemes — and how to avoid them.

✕Forgetting to raise the coefficient inside the bracket to the power
WMA12/01 examiner reports — flagged annually
Why it happens
In $(2 + 3x)^{4}$ students often compute $\binom{4}{2} \cdot 2 \cdot (3x)^{2} = 6 \cdot 2 \cdot 9x^{2}$ — instead of $\binom{4}{2} \cdot 2^{2} \cdot (3x)^{2}$ .
How to avoid it
In $\binom{n}{r} a^{n - r} b^{r}$ both $a$ AND $b$ must be raised to their respective powers, INCLUDING the numerical part of $b$ . So $(3x)^{2} = 9 x^{2}$ , and the term is $6 \cdot 4 \cdot 9 x^{2} = 216 x^{2}$ .
✕Losing the minus sign when expanding $(a - b)^{n}$
Why it happens
Students treat $(a - b)^{n}$ as $(a + b)^{n}$ and forget the alternating signs.
How to avoid it
Write $(a - b)^{n}$ as $(a + (-b))^{n}$ . Odd-power terms in $b$ get a minus sign; even-power terms keep a plus. E.g. $(2 - x)^{4} = 16 - 32x + 24 x^{2} - 8 x^{3} + x^{4}$ .
✕Setting $r = 4$ to find the term in $x^{4}$ in $(2 + x)^{6}$ — and getting the wrong term
Why it happens
Confusion between term index ( $T_{r + 1}$ ) and exponent.
How to avoid it
The exponent of $b$ equals $r$ . So in $(a + b)^{n}$ the term containing $b^{4}$ has $r = 4$ , and it's the $5$ -th term (since the first term has $r = 0$ ). When unsure, write the general term out and set the $b$ -power to the value you want.
✕Applying $\binom{n}{r}$ when $n = -2$ or $n = 1/3$
Why it happens
Confusing P2's binomial expansion (positive integer $n$ ) with P4's generalisation.
How to avoid it
In P2 (WMA12) the binomial theorem is ONLY for positive integer $n$ . The expansion is finite. Negative and fractional exponents (with their infinite series and convergence conditions) are studied in P4 (WMA14).
✕Writing the expansion in descending powers when the question says 'ascending'
Why it happens
Habit from algebra (writing polynomials in standard form).
How to avoid it
If the question says 'ascending powers of $x$ ', start with the constant and increase the exponent. Examiners may not award marks if the order is wrong, especially in 'first three terms' style questions.

Binomial expansion — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Binomial expansion

Short Notes - Binomial expansion

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Expand a binomial fully (4 marks, P2)

2First four terms of (1+kx)n(1 + kx)^{n}(1+kx)n (5 marks, P2)

3Find the coefficient of a particular term (5 marks, P2)

4Find nnn from a given coefficient (6 marks, P2)

5Approximation using a binomial (4 marks, P2)

Binomial theorem (positive integer nnn)

Binomial coefficient

Special case (1+x)n(1 + x)^{n}(1+x)n

General term (Tr+1T_{r + 1}Tr+1​)

Binomial

Binomial coefficient nCr{}^{n}C_{r}nCr​ or (nr)\binom{n}{r}(rn​)

Pascal's triangle

Factorial

Ascending powers (of xxx)

✕Forgetting to raise the coefficient inside the bracket to the power

✕Losing the minus sign when expanding (a−b)n(a - b)^{n}(a−b)n

✕Setting r=4r = 4r=4 to find the term in x4x^{4}x4 in (2+x)6(2 + x)^{6}(2+x)6 — and getting the wrong term

✕Applying (nr)\binom{n}{r}(rn​) when n=−2n = -2n=−2 or n=1/3n = 1/3n=1/3

✕Writing the expansion in descending powers when the question says 'ascending'

Binomial expansion — frequently asked questions

2First four terms of $(1 + kx)^{n}$ (5 marks, P2)

4Find $n$ from a given coefficient (6 marks, P2)

Binomial theorem (positive integer $n$ )

Special case $(1 + x)^{n}$

General term ( $T_{r + 1}$ )

Binomial coefficient ${}^{n}C_{r}$ or $\binom{n}{r}$

Ascending powers (of $x$ )

✕Losing the minus sign when expanding $(a - b)^{n}$

✕Setting $r = 4$ to find the term in $x^{4}$ in $(2 + x)^{6}$ — and getting the wrong term

✕Applying $\binom{n}{r}$ when $n = -2$ or $n = 1/3$