Binomial expansion and Pascal's triangle (QCE Math Methods Unit 1)

Syllabus dot point

“Expand expressions of the form $(a + b)^n$ for a positive integer $n$ using Pascal's triangle and binomial coefficients, and find specific terms and coefficients”

7 min readUpdated 7 Oct 2026

Quick answer

To expand $(a+b)^n$ , write the terms $\binom{n}{k}a^{n-k}b^k$ for $k=0$ to $n$ . The coefficients $\binom{n}{k}$ come from row $n$ of Pascal's triangle. For one term, use the general term formula and do not expand everything. Keep the sign and the coefficients of $a$ and $b$ inside the brackets.

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What this dot point is asking
Pascal's triangle and binomial coefficients
The binomial theorem
Expanding with a negative or a coefficient
Finding one term
Approximations
Practice questions

What this dot point is asking

In the 2025 QCAA Mathematical Methods syllabus, Unit 1 Topic 2 is titled Binomial expansion and cubic functions. For the binomial part you are expected to expand a power of a two-term expression, find a particular term or coefficient, and use the result in context. The official syllabus is at qcaa.qld.edu.au.

Pascal's triangle and binomial coefficients

Pascal's triangle starts with 1 at the top. Each entry is the sum of the two entries above it.

Row $n$	Coefficients
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

The entry in row $n$ and position $k$ (counting from $k=0$ ) is the binomial coefficient

\binom{n}{k}=\frac{n!}{k!\,(n-k)!}.

It counts the number of ways to choose $k$ items from $n$ , which is why it also appears in probability. Your calculator has an $nCr$ function, which is useful when $n$ is large.

The binomial theorem

(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k.

The powers of $a$ fall from $n$ to $0$ , and the powers of $b$ rise from $0$ to $n$ . The power of $a$ and the power of $b$ in every term add to $n$ .

Expand

(x+2)^5

Row 5 of Pascal's triangle is 1, 5, 10, 10, 5, 1. The powers of 2 are $1, 2, 4, 8, 16, 32$ .

(x+2)^5=x^5+5(2)x^4+10(4)x^3+10(8)x^2+5(16)x+32

=x^5+10x^4+40x^3+80x^2+80x+32.

Check by substituting $x=1$ . The left side is $3^5=243$ . The right side is $1+10+40+80+80+32=243$ .

Expanding with a negative or a coefficient

Treat the second term as a whole, including its sign. For $(2x-3)^4$ , take $a=2x$ and $b=-3$ and use row 4: 1, 4, 6, 4, 1.

(2x-3)^4=(2x)^4+4(2x)^3(-3)+6(2x)^2(-3)^2+4(2x)(-3)^3+(-3)^4

=16x^4-96x^3+216x^2-216x+81.

The signs alternate because $b$ is negative.

Finding one term

The general term, counted from $k=0$ , is

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

Coefficient of

x^3

(3x-2)^5

The $x^3$ term has $k=3$ factors of $3x$ and $2$ factors of $-2$ .

\binom{5}{3}(3x)^3(-2)^2=10\times 27x^3\times 4=1080x^3.

The coefficient is $1080$ .

Constant term in

\left(x+\dfrac{2}{x}\right)^6

The general term is $\binom{6}{k}x^{6-k}\left(\dfrac{2}{x}\right)^k=\binom{6}{k}2^kx^{6-2k}$ . The constant term needs $6-2k=0$ , so $k=3$ .

\binom{6}{3}2^3=20\times 8=160.

Approximations

A power of a number close to 1 can be approximated with the first few terms. Write $1.02^5=(1+0.02)^5$ .

1+5(0.02)+10(0.02)^2=1+0.1+0.004=1.104.

The calculator value is $1.1040808$ , so three terms give accuracy to about $0.0001$ .

Common trap

Three common errors. First, forgetting to raise the coefficient to the power, for example writing $(3x)^3$ as $3x^3$ . Second, dropping the negative sign when $b$ is negative, which ruins the sign pattern. Third, mixing up the term number and $k$ . The first term has $k=0$ , so the fourth term has $k=3$ .

Practice questions

1. Expand $(1+x)^4$ .

Model answer: Using row 4, $(1+x)^4=1+4x+6x^2+4x^3+x^4$ .

2. Find the coefficient of $x^2$ in $(2x+1)^5$ .

Model answer: The term with $x^2$ has $k=2$ : $\binom{5}{2}(2x)^2(1)^3=10\times 4x^2=40x^2$ . The coefficient is $40$ .

3. Use the first three terms of $(1+0.02)^5$ to approximate $1.02^5$ , then comment on the accuracy.

Model answer: The first three terms give $1+0.1+0.004=1.104$ . The exact value is $1.1040808$ , so the approximation is accurate to three decimal places, because the omitted terms are very small.

4. Show that the sum of the coefficients in the expansion of $(x+2)^5$ is $243$ .

Model answer: Substitute $x=1$ to get $(1+2)^5=3^5=243$ . This matches the sum $1+10+40+80+80+32=243$ found in the worked example.

Practise this

Sources & how we know this

Mathematical Methods General senior syllabus 2025 — QCAA