What is off-by-one with r?

The term in b^r is the (r+1)th term; the first term has r = 0.

SQA AH style: specific term-style practice: Find the term independent of x in (x^2 + 1x)^6.

General term: 6r(x^2)^6 - r(1x)^r = 6rx^12 - 2rx^-r = 6rx^12 - 3r (2 marks). Independent of x means the power is zero: 12 - 3r = 0, so r = 4 (1 mark). Term: 64 = 15, so the constant term is 15 (1 mark). Markers reward the general term, setting the power to zero, solving for r, and evaluating the coefficient.

ScotlandMathsSyllabus dot point

How do you expand a power of a binomial and pick out a specific term?

Use the binomial theorem to expand expressions of the form (a + b) to the power n for a positive integer n, using binomial coefficients, and find a general term or a specific term such as the constant term or the coefficient of a chosen power.

A focused answer to the SQA Advanced Higher Mathematics binomial theorem content, covering binomial coefficients and Pascal's triangle, the full expansion of (a + b) to the power n, the general term formula, and finding a specific term such as the constant term or the coefficient of a given power.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
Binomial coefficients
The full expansion
The general term and specific terms
Reading the structure of a term
Why it connects forward
Try this

What this dot point is asking

The SQA wants you to expand a power of a binomial, $(a + b)^n$ , for a positive integer $n$ , and to pick out individual terms without writing the whole expansion. The general-term formula is the tool that makes the second part quick.

Binomial coefficients

The coefficients $\binom{n}{r}$ can be read from Pascal's triangle, where each entry is the sum of the two above it, or computed directly from the factorial formula. For small $n$ the triangle is fastest; for larger $n$ or for a single term the formula is better.

The full expansion

To expand $(a + b)^n$ in full, run $r$ from $0$ to $n$ , taking $a^{n - r}b^{r}$ with the coefficient $\binom{n}{r}$ . Keeping track of the signs is the main source of error when $b$ is negative.

Full expansion of

(1 + 2x)^5

Expand $(1 + 2x)^5$ .

Step 1: Write the binomial coefficients

Row 5 of Pascal's triangle gives the six coefficients needed. Each entry is the sum of the two above it in the previous row:

1,\quad 5,\quad 10,\quad 10,\quad 5,\quad 1.

Step 2: Build the terms with $a = 1$ and $b = 2x$

The powers of $a = 1$ are all $1$ , so the binomial coefficients multiply the powers of $b = 2x$ directly. It is important to raise the whole bracket $2x$ to each power, not just $x$ :

1 + 5(2x) + 10(2x)^2 + 10(2x)^3 + 5(2x)^4 + (2x)^5.

Step 3: Evaluate each power and simplify

$(2x)^n = 2^n x^n$ , so multiply the numerical part from the coefficient by the power of $2$ :

= 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5.

Final answer: $(1 + 2x)^5 = 1 + 10x + 40x^2 + 80x^3 + 80x^4 + 32x^5$ .

The general term and specific terms

The single most useful formula is the general term, $T_{r+1} = \binom{n}{r}a^{n - r}b^{r}$ . To find the coefficient of $x^k$ or the term independent of $x$ , simplify the power of $x$ in the general term, set it equal to your target, solve for $r$ , and substitute back.

Finding the coefficient of

x^3

using the general term

Find the coefficient of $x^3$ in $(2 + x)^6$ .

Step 1: Write the general term

The general term of $(a + b)^n$ is $T_{r+1} = \binom{n}{r}a^{n-r}b^r$ . With $a = 2$ , $b = x$ , $n = 6$ :

T_{r+1} = \binom{6}{r} 2^{6-r} x^r,

so the power of $x$ in any term is simply $r$ .

Step 2: Match the required power

For $x^3$ we need $r = 3$ . No equation to solve here; the power of $x$ is just $r$ because $b = x$ .

Step 3: Evaluate the coefficient

Substitute $r = 3$ into the numerical part of the general term:

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

Final answer: the coefficient of $x^3$ is $160$ .

When the binomial contains a negative power of $x$ , such as $\left(x + \dfrac{1}{x}\right)^n$ , the general term has a power of $x$ that depends on $r$ in a way you must simplify before matching; the "term independent of $x$ " is then the case where that power is zero.

Reading the structure of a term

Every term in the expansion is the product of three pieces: a binomial coefficient $\binom{n}{r}$ , a power of the first quantity $a^{n - r}$ , and a power of the second quantity $b^{r}$ . When $a$ or $b$ is itself a product, such as $2x$ or $\dfrac{3}{x}$ , raising it to a power affects both the number and the variable, so $\left(\dfrac{3}{x}\right)^{r} = \dfrac{3^{r}}{x^{r}}$ contributes both a numerical factor $3^{r}$ and a negative power of $x$ . Keeping the numerical part and the power of $x$ separate is the key to simplifying the general term cleanly. Once the power of $x$ is written as a single expression in $r$ , matching it to the target power becomes a one-line equation, and the numerical factor is evaluated only at the end. This discipline, separate the number from the variable, simplify the power, then match, turns even an intimidating bracket with negative or fractional powers inside into a routine calculation.

Spotting a missing power in

(x^2 - 2)^4

Find the coefficient of $x^5$ in $(x^2 - 2)^4$ multiplied out.

Step 1: Write the general term

With $a = x^2$ , $b = -2$ , $n = 4$ , the general term is:

\binom{4}{r}(x^2)^{4-r}(-2)^r = \binom{4}{r}(-2)^r x^{8-2r}.

The power of $x$ in every term is $8 - 2r$ , which is always even.

Step 2: Match the power of $x$

For $x^5$ we need $8 - 2r = 5$ , which gives $2r = 3$ , or $r = 1.5$ . Since $r$ must be a non-negative integer, there is no such term.

Step 3: Interpret the result

Because every power of $x$ in the expansion is even (always $8 - 2r$ ), no odd power of $x$ can appear. The coefficient of $x^5$ is $0$ .

Final answer: the coefficient of $x^5$ is $0$ ; this expansion contains only even powers of $x$ .

Why it connects forward

The binomial theorem is the entry point to series work. Allowing $n$ to be a fraction or a negative number (with $|x| < 1$ ) extends it to the binomial series, an infinite expansion that overlaps with the Maclaurin series you meet next. Recognising the general-term structure here makes those infinite expansions feel familiar rather than new.

Try this

Q1. Write the coefficients for $(a + b)^4$ . [1 mark]

Cue. Row 4 of Pascal's triangle: $1, 4, 6, 4, 1$ .

Q2. Find the coefficient of $x^2$ in $(1 + 3x)^5$ . [3 marks]

Cue. $\binom{5}{2}(3)^2 = 10\times 9 = 90$ .

Exam-style practice questions

Practice questions written in the style of SQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AH style: full expansion3 marksExpand

(2x - 3)^4

using the binomial theorem.

Show worked answer →

Coefficients from row 4 of Pascal's triangle: $1, 4, 6, 4, 1$ (1 mark).

$(2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 + 4(2x)(-3)^3 + (-3)^4$ (1 mark).

$= 16x^4 - 96x^3 + 216x^2 - 216x + 81$ (1 mark). Markers reward the coefficients, the term-by-term structure with correct signs, and the simplified expansion.

AH style: specific term4 marksFind the term independent of

x

\left(x^2 + \dfrac{1}{x}\right)^6

Show worked answer →

General term: $\binom{6}{r}(x^2)^{6 - r}\left(\dfrac{1}{x}\right)^{r} = \binom{6}{r}x^{12 - 2r}x^{-r} = \binom{6}{r}x^{12 - 3r}$ (2 marks).

Independent of $x$ means the power is zero: $12 - 3r = 0$ , so $r = 4$ (1 mark).

Term: $\binom{6}{4} = 15$ , so the constant term is $15$ (1 mark). Markers reward the general term, setting the power to zero, solving for $r$ , and evaluating the coefficient.

Related dot points

Sources & how we know this

SQA Advanced Higher Mathematics Course Specification — SQA (2019)