Q: What does IP AMaths Notes (Upper Sec, Year 3-4): 10) Binomial Theorem cover? A: General term, greatest term, and approximation strategies for binomial expansions in IP AMaths.
The binomial theorem expands (a+b)n quickly. Identify the general term and know when to truncate.
Keep the full topic roadmap handy via our IP Maths tuition hub so you can jump into related drills, quizzes, or diagnostics as you move through these notes.
These tools match the SEAB GCE O-Level Additional Mathematics (4049) binomial theorem syllabus: integer-index expansions, term extraction, and fractional/negative-index series valid for ∣x∣<1
Status: SEAB O-Level Additional Mathematics 4049 syllabus (exams from 2025) checked 2025-11-30 - scope unchanged; remains the reference for this note.
The core idea is simple: Binomial theorem is a fast way to find terms without expanding everything.
Use it as a working check: Write the general term, match the required power, then substitute that r-value. For approximations, use only enough early terms for the accuracy needed.
Then go one layer deeper: Example: to find the coefficient of x^5 in (2 - x)^7, set r = 5 in the general term and keep the negative sign from (-x)^5.
Expansion Setup Map
Before choosing a formula, classify the question by the thing it asks for.
Question type
First move
What to solve for
Common trap
Coefficient of xm
Write the general term Tr+1.
The value of r that makes the power of x equal m.
Forgetting that (−x)r changes the sign when r is odd.
Constant term
Combine the powers of x inside Tr+1.
The value of r that makes the total exponent 0.
Treating the constant term as the middle term. It is not always central.
Approximation
Rewrite the number as 1+u or 1−u.
How many early terms are needed.
Using the infinite-series form when the magnitude of u is at least 1.
Greatest term
Compare neighbouring absolute terms.
Where the ratio crosses 1.
Comparing signed terms instead of magnitudes.
For coefficient questions, the structure is always:
general term→power equation→r→coefficient
Before substituting, run this index check:
Check
What it means
Mistake it prevents
r
Number of times the second term is chosen.
Calling it the term number.
Tr+1
The actual term position in the expansion.
Using Tr when the formula gives Tr+1.
an−rbr
First term power falls while second term power rises.
Keeping both powers the same way round.
Sign of br
A negative second term changes sign when r is odd.
Losing the negative sign in coefficient questions.
This is why a coefficient question should not start by guessing the term number. Solve the power equation first, then translate r into the term position only if the question asks for it.
Example: in (3x2−2x−1)5, the general term is
Tr+1=(r5)(3x2)5−r(−2x−1)r.
The power of x is 2(5−r)−r=10−3r. To find the coefficient of x1, solve 10−3r=1, so r=3. Then
T4=(35)(3x2)2(−2x−1)3=−720x.
The coefficient of x is −720.
1 Formulae to recall
General term: Tr+1=(rn)an−rbr for integer n≥0.
Greatest term near maximum when ∣Tr+1∣∣Tr+2∣≤1
For negative or fractional indices, use the infinite series form: (1+x)k=1+kx+2!k(k−1)x2+⋯
Fractional-index validity checkpoint
For negative or fractional indices, the expansion is an infinite series. Before using the first few terms, rewrite the expression into (1+u)k form and check that the small part u is between −1 and 1.
Expression
Small part u
Validity check
Answer move
(1−3x)−1/2
−3x
\(
-3x
\lt 1\), so −31<x<31.
State the interval before using the series.
(2+x)−1
Factor first: 2−1(1+2x)−1
(0.97)−1/2
−0.03
\(
-0.03
\lt 1\).
The approximation is valid because the number is close to 1.
Worked check: for (1−3x)−1/2, substituting x=0.2 gives u=−0.6, which is valid. Substituting x=0.5 gives u=−1.5, so the same binomial series about 1 is not valid.
Misconception check: checking ∣x∣<1 is not enough when the expression is (1−3x)k or (1+x/2)k. Check the full small part inside the brackets.
Approximation rewrite checkpoint
For approximation questions, the hardest step is often rewriting the number so the small part matches the series variable.
Target expression
Rewrite first
Small part to check
Common trap
(1.02)5
(1+0.02)5
0.02
Rounding too early after the first two terms.
(0.97)−1/2
(1−0.03)−1/2
−0.03
(2.06)5
25(1+0.03)5
0.03
(1−3x)−1/2
Already in (1+u)k form with u=−3x
Worked check: to approximate (2.06)5, factor out 2 first:
(2.06)5=25(1.03)5=32(1+0.03)5.
Now the expansion uses the small part 0.03, not 2.06 or 0.06.
Misconception check: the binomial series is built around 1. If the main number is not close to 1, factor out a convenient constant before expanding.
Greatest term checkpoint
For greatest-term questions, compare neighbouring terms rather than expanding the whole expression.
Step
What to do
Why it works
1
Write ∣Tr+1Tr+2∣.
This compares the next term with the current term.
2
Find where the ratio changes from greater than 1 to less than or equal to 1.
Terms increase before the crossing and decrease after it.
3
Check the neighbouring terms if the ratio equals 1.
A tie can give two equal greatest terms.
Worked check: find the greatest term in the expansion of (2+1)7. The general term is Tr+1=(r7)27−r, so
Tr+1Tr+2=r+17−r×21.
For r=1, the ratio is 26×21=1.5, so the next term is larger. For r=2, the ratio is 35×21=65, so the next term is smaller. Therefore T3 is the greatest term:
T3=(27)25=672.
Common trap: compare absolute term sizes. If the expansion has alternating signs, the greatest numerical value and the greatest magnitude may not mean the same thing.
2 Worked example - Specific term
Find the coefficient of x5 in (2−x)7.
General term: Tr+1=(r7)27−r(−x)r.
Power of x is r. Set r=5.
Substitute: T6=(57)22(−x)5=21×4×(−x5)=−84x5
Coefficient is −84.
3 Worked example - Binomial approximation
Approximate (1.02)5 to 5 significant figures without a calculator.
Write (1+0.02)5.
Use first three terms: 1+5×0.02+25×4(0.02)2=1+0.1+0.004=1.104.
