What is not combining the powers of x first?

When the bracket has both x and 1x, simplify to a single power x^n - 2r (or similar) before setting it equal to the target.

What is sign slips with a negative term?

If b is negative, the sign of b^r alternates with r; track it.

SingaporeAdditional MathematicsSyllabus dot point

How do we find one specific term, such as the term in x cubed, in a binomial expansion without writing out the whole thing?

Find a specific term or coefficient in a binomial expansion using the general term formula

A focused answer to the N(A)-Level Additional Mathematics outcome on the general term of a binomial expansion. Use the general term to pick out a specific power of x or a constant term without expanding fully.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-06

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

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to pick out a single term of a binomial expansion (for example the term in $x^3$ , or the constant term) without writing the whole expansion. The tool is the general term formula, which describes any one term in terms of its position $r$ . You set the power of $x$ equal to the one you want, solve for $r$ , and evaluate. This is faster and less error-prone than a full expansion for high powers.

The answer

The general term

In the expansion of $(a + b)^n$ , the term containing $b^r$ is:

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

where $r$ runs from $0$ (the first term) up to $n$ . Every term in the expansion is this formula for some value of $r$ , so finding a particular term is just a matter of finding the right $r$ .

Finding the term in a chosen power of x

To find the term in $x^k$ :

Write the general term with $a$ and $b$ substituted.
Combine the powers of $x$ into a single power.
Set that power equal to $k$ and solve for $r$ .
Put that $r$ back into the general term and evaluate the number.

Finding a constant (term independent of x)

A constant term is the term in $x^0$ . Combine the powers of $x$ as before, set the total power to zero, solve for $r$ , and evaluate. This is common when the bracket contains both $x$ and $\dfrac{1}{x}$ , because the powers can cancel.

Coefficient versus term

Read the question carefully. The term includes the power of $x$ (for example $15x^2$ ); the coefficient is just the number in front (here $15$ ). Give exactly what is asked.

A reliable order of working

A tidy routine prevents slips: substitute $a$ and $b$ into the general term, simplify any numerical base and combine the powers of $x$ into one, then equate that single power to the target before evaluating. Doing the index arithmetic before you reach for the calculator keeps the numbers small and makes a wrong value easy to spot.

Examples in context

Example 1. Picking a middle term. In a high-power expansion such as $(1 + x)^{10}$ , writing out all eleven terms wastes time if you only need the $x^4$ term. The general term goes straight to $\binom{10}{4}x^4 = 210x^4$ , which is the efficiency the method is designed for.

Example 2. Cancelling powers for a constant. Expressions like $\left(x^2 - \dfrac{1}{x}\right)^6$ have a constant term only when the positive and negative powers of $x$ cancel. Setting the combined power to zero finds exactly which term that is, a routine that appears often in examination questions.

Try this

Q1. Write down the general term of $(1 + x)^n$ . [1 mark]

Cue. $T_{r+1} = \binom{n}{r}x^r$ .

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

Cue. $\binom{5}{2} = 10$ , so the coefficient is $10$ .

Q3. Find the coefficient of $x^3$ in $(1 + 2x)^4$ . [3 marks]

Cue. General term $\binom{4}{r}(2x)^r$ ; for $x^3$ , $r = 3$ , giving $\binom{4}{3}2^3 = 4 \times 8 = 32$ .

Exam-style practice questions

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

Original4 marksFind the coefficient of

x^2

in the expansion of

(1 + x)^6

Show worked answer →

The general term is $\binom{6}{r}(1)^{6 - r}x^r = \binom{6}{r}x^r$ .

For the $x^2$ term, set $r = 2$ : $\binom{6}{2} = 15$ .

So the coefficient of $x^2$ is $15$ .

What markers reward: writing the general term, setting $r$ to the required power, and evaluating $\binom{6}{2} = 15$ as the coefficient.

Original4 marksFind the term independent of

x

(the constant term) in the expansion of

\left(x + \dfrac{2}{x}\right)^4

Show worked answer →

The general term is $\binom{4}{r}x^{4 - r}\left(\dfrac{2}{x}\right)^r = \binom{4}{r}2^r x^{4 - 2r}$ .

The constant term has power zero: $4 - 2r = 0$ , so $r = 2$ .

Term: $\binom{4}{2}2^2 = 6 \times 4 = 24$ .

What markers reward: forming the general term with the powers of $x$ combined to $x^{4 - 2r}$ , setting the power to zero to find $r = 2$ , and evaluating the constant as $24$ .