Binomial Theorem questions in O-Level Additional Mathematics (syllabus 4049) are some of the most predictable marks on Paper 1 and Paper 2. Almost every year, the examiner asks you to find one specific term, or the coefficient of one specific power of x, from an expansion. Once you can write down the general term confidently, these questions become almost mechanical.
Here is the exact method our tutors used when we sat these papers, plus the traps that cost students easy marks.
Start with the general term, always
For a positive integer n, the expansion of (a + b)^n has general term:
T = nCr × a^(n − r) × b^r, where r = 0, 1, 2, ..., n
This single line is the whole topic. Notice:
- The powers of a and b always add up to n.
- r counts how many times b appears, so r matches the power of b.
- nCr is read as "n choose r" and is on your calculator (usually the nCr button).
Strategy for every specific-term question:
- Write the general term with the actual a, b and n substituted in.
- Simplify the powers of x into one single index.
- Set that index equal to the power you want, and solve for r.
- Substitute r back in and evaluate.
Writing step 1 clearly is worth marks on its own, even if your arithmetic slips later.
Example 1: A straightforward coefficient
Find the coefficient of x^5 in the expansion of (2 + 3x)^8.
General term = 8Cr × 2^(8 − r) × (3x)^r = 8Cr × 2^(8 − r) × 3^r × x^r
We want x^5, so r = 5.
Term = 8C5 × 2^3 × 3^5 = 56 × 8 × 243 = 108 864
So the coefficient is 108 864.
Notice the question asked for the coefficient, not the term. The coefficient is the number only. If it had asked for the term, the answer would be 108 864x^5. Losing a mark for that is painful and completely avoidable.
Example 2: Negative and fractional powers
Find the term independent of x in the expansion of (2x − 1/x^2)^9.
"Independent of x" means the power of x is zero, in other words the constant term.
General term = 9Cr × (2x)^(9 − r) × (−1/x^2)^r
= 9Cr × 2^(9 − r) × (−1)^r × x^(9 − r) × x^(−2r)
= 9Cr × 2^(9 − r) × (−1)^r × x^(9 − 3r)
Set 9 − 3r = 0, so r = 3.
Term = 9C3 × 2^6 × (−1)^3 = 84 × 64 × (−1) = −5376
Two things to keep in mind here. First, combine the powers of x into one index before you solve, otherwise you will mismanage the negative index. Second, keep the negative sign inside the bracket so that (−1)^r is handled properly. Sign errors are the single most common reason students lose marks in this topic.
Example 3: Expansion multiplied by another bracket
Find the coefficient of x^2 in the expansion of (3 + x)(1 − 2x)^6.
Do not expand everything. Only two terms from (1 − 2x)^6 can produce x^2 after multiplication.
From (1 − 2x)^6, general term = 6Cr × (−2)^r × x^r.
- x^1 term: 6C1 × (−2) = −12
- x^2 term: 6C2 × (−2)^2 = 15 × 4 = 60
Now pair them up so the total power is x^2:
- 3 × 60x^2 = 180x^2
- x × (−12x) = −12x^2
Coefficient of x^2 = 180 − 12 = 168
Writing a short line such as "terms in x and x^2 are needed" shows the examiner your reasoning and secures method marks.
When r is not a whole number
Sometimes solving for r gives a fraction, for example r = 7/3. That means the requested power of x does not appear in the expansion at all, so the coefficient is 0. State that clearly instead of rounding r. Also check that r lies between 0 and n inclusive.
Quick checklist before you move on
- Did I write the general term with nCr, a^(n − r) and b^r?
- Are the brackets raised properly, so (3x)^r becomes 3^r x^r?
- Did I combine all powers of x into a single index before solving?
- Did the question want the term or the coefficient?
- Is my r a non-negative integer, and is the sign right?
Practise with variation, not volume
Ten near-identical questions will not prepare you for the paper. Instead, drill one of each type: a plain coefficient, a negative index, a constant term, a product of two brackets, and a question where you are given a coefficient and must find an unknown constant. That covers essentially every way MOE and Cambridge have phrased this in recent years.
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