What does this module cover in N(A)-Level A-Maths 4051?

It covers three outcomes in the Algebra strand of syllabus 4051: the binomial theorem for expanding $(a + b)^n$ for a positive integer $n$; finding a particular term or coefficient using the general term; and expressing a proper algebraic fraction with linear factors as partial fractions. The first two are linked methods; partial fractions is a separate algebraic skill.

What is a binomial coefficient $\binom{n}{r}$?

The binomial coefficient $\binom{n}{r}$, also written $^{n}C_{r}$, counts the ways of choosing $r$ items from $n$ and equals $\dfrac{n!}{r!(n-r)!}$. In the binomial theorem it is the coefficient of the term $a^{n-r}b^{r}$. The same numbers appear as the rows of Pascal's triangle.

How do you find one specific term without expanding the whole bracket?

Use the general term $\binom{n}{r} a^{n-r} b^{r}$. Decide which power of the variable you want, set the index on that variable equal to it to find $r$, then substitute that $r$ back to get the term or its coefficient. This avoids writing out the full expansion.

When can a fraction be split into partial fractions?

At N(A)-Level the fraction must be proper, meaning the degree of the numerator is less than the degree of the denominator, and the denominator must factorise into distinct linear factors. Each linear factor $(ax + b)$ then contributes a term $\dfrac{A}{ax + b}$, and you solve for the unknown constants.

How is the Algebra strand assessed in syllabus 4051?

Both Paper 1 and Paper 2 (each 1 hour 45 minutes, 70 marks, weighted 50 percent) draw on the whole syllabus, and candidates answer all questions. Binomial and partial-fraction questions usually appear as short, self-contained items, so a reliable method earns full marks quickly.

SingaporeAdditional Mathematics

The binomial theorem and partial fractions in N(A)-Level Additional Mathematics (4051): expanding a power of a bracket, finding a particular term, and splitting a proper algebraic fraction into simpler parts

An N(A)-Level Additional Mathematics (4051) overview of the binomial theorem and partial fractions in the Algebra strand. How to expand a power of a bracket using binomial coefficients, how the general term picks out a specific term or coefficient, and how to split a proper algebraic fraction with linear factors, with links to every dot point.

Generated by Claude Opus 4.88 min readSEAB-4051 AlgebraUpdated 2026-06-13

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

Jump to a section

Two ways to handle brackets and fractions
The binomial theorem
Finding a particular term
Partial fractions
How this module is examined
Check your knowledge

Two ways to handle brackets and fractions

This module pairs two algebraic techniques that the N(A)-Level Additional Mathematics syllabus (SEAB 4051) treats together. The binomial theorem expands a power of a bracket quickly and lets you reach a single term without the full expansion. Partial fractions runs the other way, breaking one fraction into a sum of simpler ones. Both reward a clear, repeatable method.

This overview links three dot points: the binomial theorem itself, finding a particular term, and partial fractions. Each has its own worked answers and practice.

The binomial theorem

The binomial theorem outcome expands $(a + b)^n$ for a positive integer $n$ as

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r},

where the coefficients

\binom{n}{r}

are the binomial coefficients, also the rows of Pascal's triangle. For small

n

, reading the coefficients off Pascal's triangle is fast; for larger

n

, the

\binom{n}{r}

formula is more reliable.

Finding a particular term

The finding a particular term outcome uses the general term $\binom{n}{r} a^{n-r} b^{r}$ to pick out a single term. You write the general term, set the power of the variable you want equal to the target, solve for $r$ , and substitute back. This is much faster than expanding fully when only one coefficient is needed.

Partial fractions

The partial fractions outcome splits a proper fraction whose denominator is a product of distinct linear factors. Each factor $(ax + b)$ gives a term $\dfrac{A}{ax + b}$ . You combine over a common denominator, equate numerators, and find the constants by substituting convenient values of $x$ (often the roots of the factors).

How this module is examined

Both papers, all questions. Paper 1 and Paper 2 (each 1 hour 45 minutes, 70 marks, 50 percent) cover the full syllabus, and you answer every question.
State the general term. For a particular-term question, write $\binom{n}{r} a^{n-r} b^{r}$ before substituting; it shows the method and protects the marks.
Check the fraction is proper first. Partial fractions at this level assume a proper fraction with distinct linear factors, so confirm the numerator's degree is lower than the denominator's before splitting.

Check your knowledge

Short questions across the three outcomes. Work them with full method, then check the solutions.

Expand $(1 + x)^4$ in ascending powers of $x$ . (3 marks)
Find the coefficient of $x^3$ in $(1 + x)^5$ . (2 marks)
Evaluate $\binom{7}{2}$ . (1 mark)
Express $\dfrac{1}{(x)(x + 1)}$ in partial fractions. (3 marks)

Solutions

The four questions cover the binomial theorem, finding a particular term, evaluating a binomial coefficient, and splitting a fraction into partial fractions.

Step 1: Expand $(1 + x)^4$ using Pascal's triangle coefficients

The fourth row of Pascal's triangle gives the coefficients $1, 4, 6, 4, 1$ . Each term has the form $\binom{4}{r} x^r$ with powers of $x$ rising from $0$ to $4$ , so writing them out in ascending order gives:

(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4.

Final answer (Q1): $(1 + x)^4 = 1 + 4x + 6x^2 + 4x^3 + x^4$ .

Step 2: Find the coefficient of $x^3$ in $(1 + x)^5$

The general term is $\binom{5}{r}(1)^{5-r}(x)^r = \binom{5}{r} x^r$ . To get the $x^3$ term, set $r = 3$ and evaluate the coefficient:

\binom{5}{3} = \frac{5!}{3!\,2!} = \frac{5 \times 4}{2} = 10.

Final answer (Q2): The coefficient of $x^3$ is $10$ .

Step 3: Evaluate $\binom{7}{2}$ using the formula

The binomial coefficient formula is $\binom{n}{r} = \dfrac{n!}{r!(n-r)!}$ . Substituting $n = 7$ and $r = 2$ cancels most of the factorial:

\binom{7}{2} = \frac{7!}{2!\,5!} = \frac{7 \times 6}{2 \times 1} = 21.

Final answer (Q3): $\binom{7}{2} = 21$ .

Step 4: Express $\dfrac{1}{x(x+1)}$ in partial fractions

Write the fraction in the assumed form with one unknown constant per linear factor, then multiply both sides by the common denominator:

\frac{1}{x(x + 1)} = \frac{A}{x} + \frac{B}{x + 1} \implies 1 = A(x + 1) + Bx.

Substituting $x = 0$ eliminates $B$ and gives $A = 1$ . Substituting $x = -1$ eliminates $A$ and gives $B = -1$ .

\frac{1}{x(x+1)} = \frac{1}{x} - \frac{1}{x + 1}.

Final answer (Q4): $\dfrac{1}{x} - \dfrac{1}{x + 1}$ .

Sources & how we know this

Singapore-Cambridge GCE N(A)-Level Additional Mathematics (Syllabus 4051) — Singapore Examinations and Assessment Board (2026)