IP AMaths Notes (Upper Sec, Year 3-4): 10) Binomial Theorem

For Integrated Programme students: Your current school materials, teacher instructions, and assessment scope take precedence because IP topic sequence and depth vary by school. This is an Eclat IP guide, not the O-Level / SEC G3 exam-track guide.

How this chapter applies

• Eclat core: positive-integer binomial expansions, factorial notation, combinations, and the general term form the K341-linked core.
• School-sensitive extension: greatest-term questions, negative or fractional powers, infinite series, and convergence are outside K341 and depend on the school.
• 2027 national comparison: K341 Topic A5 requires the Binomial Theorem only for positive integer n and explicitly states that greatest-term and coefficient-property knowledge is not required.
• Check your school: confirm whether non-integer expansions are taught for IP enrichment or later pathway preparation.
• Exam-track route: use the separate O-Level and SEC G3 Additional Mathematics notes for K341 topic ownership.

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.

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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.

For coefficient questions, the structure is always:

$$\text{general term} \rightarrow \text{power equation} \rightarrow r \rightarrow \text{coefficient}$$

Before substituting, run this index check:

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 $(3x^2 - 2x^{-1})^5$, the general term is

$$T_{r+1} = \binom{5}{r}(3x^2)^{5-r}(-2x^{-1})^{r}.$$

The power of $x$ is $2(5-r) - r = 10 - 3r$. To find the coefficient of $x^1$, solve $10 - 3r = 1$, so $r = 3$. Then

$$T_4 = \binom{5}{3}(3x^2)^2(-2x^{-1})^{3} = -720x.$$

The coefficient of $x$ is $-720$.

1 Formulae to recall

• General term: $T_{r+1} = \binom{n}{r} a^{n-r} b^{r}$ for integer $n \geq 0$.
• Greatest term near maximum when $\dfrac{\left|T_{r+2}\right|}{\left|T_{r+1}\right|} \leq 1$.
• For negative or fractional indices, use the infinite series form: $(1 + x)^k = 1 + kx + \dfrac{k(k - 1)}{2!} x^2 + \cdots$ for $|x| \lt 1$.

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 $1$.

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 $|x| \lt 1$ 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.

Worked check: to approximate $(2.06)^5$, factor out 2 first:

$$(2.06)^5 = 2^5(1.03)^5 = 32(1+0.03)^5.$$

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.

Worked check: find the greatest term in the expansion of $(2 + 1)^7$. The general term is $T_{r+1} = \binom{7}{r}2^{7-r}$, so

$$\frac{T_{r+2}}{T_{r+1}} = \frac{7-r}{r+1}\times \frac{1}{2}.$$

For $r = 1$, the ratio is $\dfrac{6}{2}\times\dfrac{1}{2} = 1.5$, so the next term is larger. For $r = 2$, the ratio is $\dfrac{5}{3}\times\dfrac{1}{2} = \dfrac{5}{6}$, so the next term is smaller. Therefore $T_3$ is the greatest term:

$$T_3 = \binom{7}{2}2^5 = 672.$$

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 $x^5$ in $(2 - x)^7$.

1. General term: $T_{r+1} = \binom{7}{r} 2^{7-r} (-x)^{r}$.
2. Power of $x$ is $r$. Set $r = 5$.
3. Substitute: $T_6 = \binom{7}{5} 2^{2} (-x)^{5} = 21 \times 4 \times (-x^{5}) = -84 x^5$.

Coefficient is $-84$.

3 Worked example - Binomial approximation

Approximate $(1.02)^5$ to 5 significant figures without a calculator.

1. Write $(1 + 0.02)^5$.
2. Use first three terms: $1 + 5 \times 0.02 + \frac{5 \times 4}{2} (0.02)^{2} = 1 + 0.1 + 0.004 = 1.104$.
3. Third-order term: $\frac{5 \times 4 \times 3}{6} (0.02)^{3} = 10 \times 8 \times 10^{-6} = 8.0 \times 10^{-5}$.
4. Sum: $1.10408$, which rounds to $1.1041$ (5 s.f.).

4 Worked example - Constant term in a rational power expansion

Find the constant term in the expansion of $(x^2 - x^{-1})^6$.

1. General term: $T_{r+1} = C(6, r) \times (x^2)^{6 - r} \times (-x^{-1})^r$.
2. For a constant term, set the exponent of $x$ to zero: $12 - 3 r = 0$, so $r = 4$.
3. Substitute $r = 4: T_5 = C(6, 4) \times (-1)^4 \times x^0 = 15$.
4. Neighbouring $r$ values give non-zero exponents (they change in steps of $3$), so no other constant appears.

Answer: the constant term equals $15$.

5 Practice Quiz

Drill general term selection, coefficient extraction, and convergence-aware approximations before branching into fractional powers.

6 Try this

Determine the interval of convergence for $(1 - 3x)^{-1/2}$ and use the first three terms to approximate $(0.97)^{-1/2}$.

Endpoint check for the final exercise: at $x=-1/3$, the series is alternating with terms tending to zero, so it converges. At $x=1/3$, all terms are positive and the series diverges. The full interval is therefore $-1/3 \leq x \lt 1/3$; the strict interior condition used above guarantees absolute convergence.