7. Binomial Theorem – Class 11 Mathematics

Introduction

The Binomial Theorem is one of the most significant results in algebra. It provides a simple way to expand expressions of the type (a + b)ⁿ, where n is a positive integer.
Instead of multiplying the binomial term repeatedly, the theorem gives a direct formula that saves both time and effort.

• The word binomial means an expression with two terms, such as (x + y), (2a – 3b), (p + q).
• The theorem tells us how to expand the powers of such binomials systematically.
• It has applications in algebra, probability, statistics, calculus, and beyond.

This chapter focuses on Binomial Theorem for positive integral indices, its proprties, coefficients, middle terms, and uses in competitive exams like JEE, KCET, and COMEDK.

Historical Background

The idea of expanding binomials has existed since ancient times:

• Indian mathematicians like Brahmagupta (7th century) and Bhaskara II (12th century) explored early forms of binomial expansion.
• Chinese mathematician Jia Xian (11th century) developed methods that later formed Pascal’s triangle.
• Sir Isaac Newton extended the theorem in the 17th century to rational and real exponents (Newton’s Generalized Binomial Theorem).

Statement of the Theorem

For a positive integer n,

(a + b)ⁿ = Σ (nCr) a(n–r) br, where r = 0, 1, 2, …, n.

Expanded form:

(a + b)ⁿ = nC0 aⁿ + nC1 a(n–1)b + nC2 a(n–2)b² + … + nCr a(n–r)br + … + nCn bⁿ

Example: Expand (x + y)⁵

(x + y)⁵ = 5C0 x⁵ + 5C1 x⁴y + 5C2 x³y² + 5C3 x²y³ + 5C4 xy⁴ + 5C5 y⁵
= x⁵ + 5x⁴y + 10x³y² + 10x²y³ + 5xy⁴ + y⁵

Structure of the Expansion

1. Number of Terms: Always n + 1.
2. Powers of a: Start from n and decrease to 0.
3. Powers of b: Start from 0 and increase to n.
4. Coefficients: Given by nCr = n! / [r!(n – r)!].

Properties of Binomial Coefficients

1. nC0 = nCn = 1
2. nCr = nC(n – r) (symmetry property)
3. Recursive identity: nCr + nC(r – 1) = (n+1)Cr → basis of Pascal’s triangle.

Pascal’s Triangle

        1

       1   1

     1   2   1

   1   3   3   1

 1   4   6   4   1

Each row represents binomial coefficients.

Middle Term(s) in Expansion

• If n is even → only 1 middle term: (n/2 + 1)th term.
• If n is odd → 2 middle terms: ((n+1)/2)th and ((n+3)/2)th terms.

General Term

The (r+1)th term in (a + b)ⁿ is:
T(r+1) = nCr a(n–r) br

Special Cases

1. (a + b)² = a² + 2ab + b²
2. (a – b)² = a² – 2ab + b²
3. (a + b)³ = a³ + 3a²b + 3ab² + b³
4. (a – b)³ = a³ – 3a²b + 3ab² – b³

Applications

1. Expansion of higher powers: e.g., (x + 1)¹⁰.
2. Finding coefficients in polynomials.
3. Finding specific terms like the 7th term.
4. Approximations for small values of x (important in physics).
5. Probability: In binomial distribution (nCr pr q(n–r)).

Solved Examples

Example 1: Expand (x + y)⁴

(x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Example 2: Find the 5th term of (2x – 3)⁶

General term = nCr a(n–r) br
Here n = 6, a = 2x, b = –3
T5 = 6C4 (2x)² (–3)⁴
= 15 × 4x² × 81
= 4860x²

Example 3: Find coefficient of x⁵ in (1 + x)¹²

Required term = 12C5 x⁵
Coefficient = 12C5 = 792

Example 4: Sum of coefficients in (x – 2)⁷

Put x = 1 → (1 – 2)⁷ = (–1)⁷ = –1

Example 5: Middle term in (x + y)⁶

n = 6 → middle term = (6/2 + 1)th = 4th term
T4 = 6C3 x³ y³ = 20x³y³

Example 6: JEE Style

Find constant term in (x² + 1/x)⁵.
General term = 5Cr (x²)(5–r) (1/x)r = 5Cr x(10–2r – r) = 5Cr x(10 – 3r)
For constant term: 10 – 3r = 0 → r = 10/3 (not integer) → No constant term

Example 7: KCET Style

Find coefficient of x³ in (1 + 2x)⁶.
Term containing x³ → r = 3
Coefficient = 6C3 (2)³ = 20 × 8 = 160

Example 8: Approximation

Approximate (1.01)⁵ up to two decimals.
(1 + 0.01)⁵ = 1 + 5(0.01) + 10(0.01)² + … ≈ 1 + 0.05 + 0.001 = 1.051

Marks Distribution

• CBSE / State Boards: 6–8 marks
• JEE Main: 1 question (4 marks) almost every year
• KCET / COMEDK: 1–2 questions (direct application)
• JEE Advanced: Identity and combinatorial proofs

Common Pitfalls

1. Mixing signs when b is negative.
2. Confusing “r” with “term number”.
3. Forgetting factorial simplifications.
4. Ignoring domain restrictions in approximations.

Practice Problems

1. Expand (a – b)⁵.
2. Find the 7th term in (2x + 3)¹².
3. Find coefficient of x⁴ in (x² – 1/x)¹⁰.
4. Find middle term(s) in (x – 2)⁹.
5. Prove that sum of coefficients in (1 – 1)ⁿ = 0.

FAQs

Q1. What is the general term in Binomial Theorem?
T(r+1) = nCr a(n–r) br

Q2. How to find the middle term?

• If n is even → (n/2 + 1)th term
  
• If n is odd → ((n+1)/2)th and ((n+3)/2)th terms

Q3. What is the importance of Binomial Theorem in JEE?
Direct expansions, coefficients, and approximations are frequently tested.

Q4. Can Binomial Theorem be used for negative exponents?
Yes, through Newton’s Generalized Binomial Theorem (not part of Class 11 syllabus).

Final Summary

• The Binomial Theorem expands (a + b)ⁿ without lengthy multiplication.
• Coefficients are given by combinations (nCr).
• Applications include expansions, coefficient finding, probability, and approximations.
• Highly important for both Board Exams and Competitive Exams (JEE/KCET/COMEDK).