Binomial Theorem

IB Mathematics Analysis and Approaches HL· Topic 1: Number & Algebra· 5 min read

1. Binomial Theorem Basics and Binomial Coefficients★★☆☆☆⏱ 15 min

📘 Definition

Binomial Theorem

(a + b)^n, n \in \mathbb{Z}^+

For any positive integer $n$ , the expansion of $(a + b)^{n}$ is the sum of terms of the form $(k n) a^{n - k} b^{k}$ for $k$ from $0$ to $n$ .

Example:

For $n = 2$ , $(a + b)^{2} = a^{2} + 2 ab + b^{2}$

📐 Worked Example

Calculate $(2 5)$ using both the combination formula and Pascal's triangle.

1
Recall the combination formula for binomial coefficients:
2
$\binom{n}{k} = \frac{n!}{k!(n-k)!}$
3
Substitute $n = 5, k = 2$ :
4
$\binom{5}{2} = \frac{5!}{2! 3!} = \frac{120}{2 \times 6} = 10$
5
For Pascal's triangle, row 5 (starting at $n = 0$ ) is $[1, 5, 10, 10, 5, 1]$ . The entry at index 2 is 10, matching the formula result.

Exam tip:

Always check that the sum of the exponents of $a$ and $b$ in any term equals $n$ for a quick error check.

2. Expanding Full Binomial Expressions★★★☆☆⏱ 20 min

For small $n$ , Pascal's triangle gives quick coefficients, while for larger $n$ , the combination formula is more reliable. Always include the sign of any negative term when expanding.

📐 Worked Example

Expand $(2 x - 3 y)^{4}$ fully.

1
Identify $a = 2 x$ , $b = - 3 y$ , $n = 4$ . Binomial coefficients for $n = 4$ are $1, 4, 6, 4, 1$ .
2
Calculate each term individually:
3
$k=0: 1(2x)^4(-3y)^0 = 16x^4$
4
$k=1: 4(2x)^3(-3y)^1 = -96x^3y$
5
$k=2: 6(2x)^2(-3y)^2 = 216x^2y^2$
6
$k=3: 4(2x)^1(-3y)^3 = -216xy^3$
7
$k=4: 1(2x)^0(-3y)^4 = 81y^4$
8
Combine all terms to get the full expansion:
9
$(2x - 3y)^4 = 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$

✓ Quick check

What is the coefficient of $x^{3}$ in $(x + 2)^{5}$ ?

What is the coefficient of $x^{3}$ in $(x + 2)^{5}$ ?
- 10
- 40
- 80
- 20
Reveal answer
40 —
The term is $(3 5) x^{3} (2)^{2} = 10 \times 4 x^{3} = 40 x^{3}$ , so the coefficient is 40.

3. Finding a Specific Term or Coefficient★★★☆☆⏱ 20 min

Most IB exam questions do not require a full expansion. Instead, they ask for the coefficient of a specific power of $x$ , or the constant term, which we can find using the general term formula.

📘 Definition

General Term

The $(k + 1)$ th term in the expansion of $(a + b)^{n}$ is $T_{k + 1} = (k n) a^{n - k} b^{k}$ , where $k$ ranges from $0$ to $n$ .

📐 Worked Example

Find the constant term in the expansion of $(x^{2} + \frac{2}{x})^{6}$ .

1
Write the general term for this expansion:
2
$T_{k+1} = \binom{6}{k} (x^2)^{6 - k} \left(\frac{2}{x}\right)^k$
3
Simplify the exponent of $x$ :
4
$\text{exponent} = 2(6 - k) - k = 12 - 3k$
5
We want the constant term, so set the exponent of $x$ equal to 0:
6
$12 - 3k = 0 \implies k = 4$
7
Substitute $k = 4$ back to calculate the constant term:
8
$T_5 = \binom{6}{4} (x^2)^2 \left(\frac{2}{x}\right)^4 = 15 \times x^4 \times \frac{16}{x^4} = 240$

Exam tip:

Term number is $k + 1$ , not $k$ . If asked for the 5th term, use $k = 4$ , not $k = 5$ .

4. Approximation Applications★★★★☆⏱ 15 min

The binomial theorem is commonly used to approximate values of the form $(1 + x)^{n}$ when $x$ is very small, so higher power terms become negligible.

📐 Worked Example

Approximate $(1.02)^{8}$ correct to 3 decimal places using the binomial theorem.

1
Rewrite $1.02$ as $1 + 0.02$ , so we expand $(1 + 0.02)^{8}$ :
2
$k=0: 1(1)^8(0.02)^0 = 1$
3
$k=1: 8(1)^7(0.02)^1 = 0.16$
4
$k=2: 28(1)^6(0.02)^2 = 0.0112$
5
$k=3: 56(1)^5(0.02)^3 = 0.000448$
6
Terms for $k \geq 4$ are smaller than 0.000001, negligible for 3 decimal place accuracy. Sum the terms:
7
$1 + 0.16 + 0.0112 + 0.000448 = 1.171648 \approx 1.172$

5. Common Pitfalls

Wrong move:

Forgetting to include the negative sign when expanding $(a - b)^{n}$ , treating $b$ as positive.

Why:

This leads to incorrect signs on alternating terms in the expansion.

Correct move:

Always write $b$ as $- ∣ b ∣$ , so $(- ∣ b ∣)^{k}$ gives the correct sign automatically.

Wrong move:

Taking $k = 5$ for the 5th term, instead of $k = 4$ .

Why:

The general term starts at $k = 0$ for the first term, not $k = 1$ .

Correct move:

Remember $T_{k + 1}$ is the $(k + 1)$ th term: subtract 1 from the term number to get $k$ .

Wrong move:

Incorrect exponent simplification, e.g. $(x^{2})^{n - k} = x^{2 + n - k}$ instead of $x^{2 (n - k)}$ .

Why:

Misapplication of the exponent power rule leads to wrong $k$ and wrong coefficient.

Correct move:

Always write out the exponent simplification step by step, use $(x^{m})^{n} = x^{mn}$ .

Wrong move:

Using the finite binomial formula for non-positive integer $n$ .

Why:

Confusing the positive integer binomial theorem with the infinite general binomial series.

Correct move:

Only use the sum from $k = 0$ to $n$ when $n$ is a positive integer, use the general series for other exponents.

6. Quick Reference Cheatsheet

Concept	Formula / Rule
Binomial Theorem	$(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k}, n \in Z^{+}$
Binomial Coefficient	$(k n) = \frac{n !}{k ! ( n - k )!}, (0 n) = 1$
General Term	$T_{k + 1} = (k n) a^{n - k} b^{k}$ , $k = 0$ → first term
Constant Term	Set exponent of $x$ = 0, solve for $k$

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2025 · 1
Find coefficient of x^5
2024 · 2
Approximate (1.03)^7
2023 · 1
Find constant term in expansion

Going deeper

syllabusIB AA HL Official SyllabusOfficial reference for Topic 1

What's Next

Mastering the binomial theorem for positive integer exponents is foundational for combinatorics, probability, and the general binomial series for non-integer exponents, which is used for approximation problems in calculus and series. This topic is frequently combined with combinations to solve counting problems in IB exams. Building on this knowledge, you can explore further topics in algebra and series that build on core binomial concepts.