MathematicsUnit 28 min read

Binomial Theorem: Expansion, Pascal’s Triangle, Applications

Unit 2 of Mathematics explains the Binomial Theorem, how to expand expressions like $(a + b)^n$, Pascal’s Triangle, and its real-world uses in probability, algebra, and calculus—with step-by-step examples and NEB-style questions.

Padhai · Based on the official syllabus and past papers
Updated 28 Sept 2026

What is the Binomial Theorem?

The Binomial Theorem helps us expand expressions of the form $(a + b)^{n}$ where $n$ is a positive integer. Instead of multiplying $(a + b)$ by itself $n$ times, we use a formula to write the expansion quickly.

Key Idea:

For any positive integer $n$ , $(a + b)^{n} = k = 0 \sum n (k n) a^{n - k} b^{k}$ where $(k n)$ (read as "n choose k") is the binomial coefficient.

1. Binomial Coefficients and Pascal’s Triangle

The coefficients in the expansion are called binomial coefficients. They can be found using factorials or Pascal’s Triangle.

Factorial Notation:

$n! = n \times (n - 1) \times (n - 2) \times \dots \times 1$ For example, $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$ .

Binomial Coefficient Formula:

$(k n) = \frac{n !}{k ! ( n - k )!}$ Example: Find $(2 5)$ . $(2 5) = \frac{5 !}{2 ! ( 5 - 2 )!} = \frac{120}{2 \times 6} = 10$

Pascal’s Triangle:

Pascal’s Triangle is a triangular array where each number is the sum of the two directly above it. The $n^{t h}$ row gives the coefficients for $(a + b)^{n}$ .

Example: Use Pascal’s Triangle to expand $(x + y)^{3}$ . From Row 3: Coefficients are 1, 3, 3, 1. So, $(x + y)^{3} = 1 x^{3} + 3 x^{2} y + 3 x y^{2} + 1 y^{3}$

2. Binomial Expansion Formula

The general expansion is: $(a + b)^{n} = k = 0 \sum n (k n) a^{n - k} b^{k}$ This means:

The first term is $a^{n}$ (when $k = 0$ ).
The last term is $b^{n}$ (when $k = n$ ).
The powers of $a$ decrease while the powers of $b$ increase.

Graphical representation of the expansion of (2x − 3y)⁴, showing each term's contribution.

Worked Example 1:

Expand $(2 x - 3 y)^{4}$ . Solution: Here, $a = 2 x$ , $b = - 3 y$ , and $n = 4$ . Using the formula: $(2 x - 3 y)^{4} = k = 0 \sum 4 (k 4) (2 x)^{4 - k} (- 3 y)^{k}$ Calculate each term:

$k = 0$ : $(0 4) (2 x)^{4} (- 3 y)^{0} = 1 \times 16 x^{4} \times 1 = 16 x^{4}$
$k = 1$ : $(1 4) (2 x)^{3} (- 3 y)^{1} = 4 \times 8 x^{3} \times (- 3 y) = - 96 x^{3} y$
$k = 2$ : $(2 4) (2 x)^{2} (- 3 y)^{2} = 6 \times 4 x^{2} \times 9 y^{2} = 216 x^{2} y^{2}$
$k = 3$ : $(3 4) (2 x)^{1} (- 3 y)^{3} = 4 \times 2 x \times (- 27 y^{3}) = - 216 x y^{3}$
$k = 4$ : $(4 4) (2 x)^{0} (- 3 y)^{4} = 1 \times 1 \times 81 y^{4} = 81 y^{4}$

Final Expansion: $(2 x - 3 y)^{4} = 16 x^{4} - 96 x^{3} y + 216 x^{2} y^{2} - 216 x y^{3} + 81 y^{4}$

3. Special Cases and Shortcuts

Identifying the middle term (k=3) in the expansion of (3x + 2y)⁶ (n=6, even).

Case 1: $(1 + x)^{n}$

When $a = 1$ and $b = x$ , the expansion simplifies to: $(1 + x)^{n} = k = 0 \sum n (k n) x^{k}$ Example: Expand $(1 + 2 x)^{3}$ . Using $a = 1$ , $b = 2 x$ , $n = 3$ : $(1 + 2 x)^{3} = 1 + 3 (2 x) + 3 (2 x)^{2} + (2 x)^{3} = 1 + 6 x + 12 x^{2} + 8 x^{3}$

Graphical comparison of (1 + x)³ and its expansion

Case 2: $(a - b)^{n}$

When $b$ is negative, the signs alternate: $(a - b)^{n} = k = 0 \sum n (k n) a^{n - k} (- b)^{k}$ Example: Expand $(x - 1)^{4}$ . $(x - 1)^{4} = x^{4} - 4 x^{3} + 6 x^{2} - 4 x + 1$

Case 3: Middle Term

For even $n$ , the middle term is the $\frac{n}{2} + 1^{t h}$ term. For odd $n$ , there are two middle terms: $\frac{n + 1}{2}^{t h}$ and $\frac{n + 3}{2}^{t h}$ .

Identifying the middle term in binomial expansion (n=6)

Example: Find the middle term of $(3 x + 2 y)^{6}$ . Since $n = 6$ (even), the middle term is the $4^{t h}$ term ( $k = 3$ ): $(3 6) (3 x)^{3} (2 y)^{3} = 20 \times 27 x^{3} \times 8 y^{3} = 4320 x^{3} y^{3}$

4. Applications of Binomial Theorem

Comparison of exact and approximate values of (1.01)⁵ using binomial approximation.

1. Algebraic Expansions

Used to simplify expressions like $(x + y)^{n}$ quickly.

2. Probability

In probability, binomial coefficients count the number of ways to get $k$ successes in $n$ trials (e.g., coin tosses).

3. Approximations

For small $x$ , $(1 + x)^{n} \approx 1 + n x$ (first two terms of expansion).

Example: Approximate $(1.01)^{5}$ . Let $x = 0.01$ , $n = 5$ : $(1 + 0.01)^{5} \approx 1 + 5 (0.01) = 1.05$ Actual value: $1.05101005$ (close approximation).

5. Comparison Table: Factorial vs. Binomial Coefficients

Feature	Factorial ( $n!$ )	Binomial Coefficient ( $(k n)$ )
Definition	Product of first $n$ natural numbers.	Number of ways to choose $k$ items from $n$ .
Formula	$n! = n \times (n - 1) \times \dots \times 1$	$\frac{n !}{k ! ( n - k )!}$
Example	$4! = 24$	$(2 4) = 6$
Use in Expansion	Used to compute coefficients.	Directly gives coefficients in $(a + b)^{n}$ .

6. Common Mistakes to Avoid

Incorrect Signs: Forgetting to include negative signs when $b$ is negative (e.g., $(a - b)^{n}$ ).
Wrong Powers: Mixing up the powers of $a$ and $b$ (e.g., writing $a^{k} b^{n - k}$ instead of $a^{n - k} b^{k}$ ).
Factorial Errors: Misapplying factorial rules (e.g., $(k n) \neq = \frac{n}{k}$ ).
Pascal’s Triangle Misuse: Using the wrong row for the exponent $n$ .

Exam Tip

Memorize the Formula: $(a + b)^{n} = k = 0 \sum n (k n) a^{n - k} b^{k}$ Know how to apply it for $(a - b)^{n}$ and $(1 + x)^{n}$ .
Practice Expansions: NEB often asks for full expansions (e.g., $(2 x + 3 y)^{4}$ ) or specific terms (e.g., the $5^{t h}$ term of $(x - 2)^{6}$ ).
Use Pascal’s Triangle for Small $n$ : For $n \leq 5$ , Pascal’s Triangle is faster than factorials.
Watch for Tricks:
- If the question asks for the general term, use $T_{k + 1} = (k n) a^{n - k} b^{k}$ .
- For approximations, keep only the first two terms if $x$ is small.
Check Your Work:
- Verify the first and last terms (should be $a^{n}$ and $b^{n}$ ).
- Ensure signs alternate correctly for $(a - b)^{n}$ .

NEB Board-Style Questions

Short Answer (5 marks each)

Expand $(3 x - 2 y)^{4}$ using the Binomial Theorem.
Find the $4^{t h}$ term in the expansion of $(\frac{x}{2} + \frac{2}{x})^{6}$ .
Using Pascal’s Triangle, write the expansion of $(a + b)^{5}$ .
Approximate $(0.99)^{4}$ using the first two terms of the Binomial expansion.
Find the middle term(s) of $(2 x + 3 y)^{6}$ .

Long Answer (10 marks)

Prove that the sum of the coefficients in the expansion of $(1 + x + x^{2})^{n}$ is $3^{n}$ .
A box contains 5 red and 3 blue balls. Three balls are drawn at random. Find the probability that exactly 2 are red using the Binomial Theorem.
Expand $(x + \frac{1}{x})^{5}$ and hence find the value of $(1 + \frac{1}{2})^{5} + (1 - \frac{1}{2})^{5}$ .

Objective (1 mark each)

The coefficient of $x^{3}$ in $(2 x + 1)^{5}$ is: a) 40 b) 80 c) 10 d) 20
The general term in the expansion of $(x^{2} - y)^{6}$ is: a) $(k 6) x^{12 - 2 k} (- y)^{k}$ b) $(k 6) x^{2 k} (- y)^{6 - k}$ c) $(k 6) x^{6 - k} (- y)^{2 k}$ d) $(k 6) x^{12 - k} (- y)^{k}$

Answers:

$81 x^{4} - 216 x^{3} y + 216 x^{2} y^{2} - 96 x y^{3} + 16 y^{4}$
$60 x^{2}$
$a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5}$
$0.9606$ (approx.)
$640 x^{3} y^{3}$
[Proof involves setting $x = 1$ and using the expansion.]
$\frac{( 2 5 ) ( 1 3 )}{( 3 8 )} = \frac{30}{56} = \frac{15}{28}$
$32 + \frac{1}{32}$
a) 40
a) $(k 6) x^{12 - 2 k} (- y)^{k}$

Based on the NEB +2 Science syllabus for Mathematics (Maths), unit 2.

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