NCERT Solutions for Class 11 Maths Chapter 8 – Binomial Theorem
Long multiplication is tedious. Try expanding (x + y)⁸ by hand, and you will spend ten minutes and probably make an error somewhere in the middle. The Binomial Theorem solves this completely: it gives you a direct formula to find any term in the expansion of (x + y)ⁿ without ever multiplying it out step by step.
This is one of those chapters where the payoff is immediate and satisfying. Learn the theorem, understand the general term formula, and you can find the coefficient of x⁵ in (2x + 3)⁸ in under two minutes — something that would otherwise require writing out the entire expansion. NCERT's Chapter 8 develops this theorem from first principles using mathematical induction (the technique from Chapter 4), then explores its properties and applications in depth. For all Chapters must, read NCERT Solutions for Class 11 Maths and subject-wise NCERT Solutions for Class 11.
The NCERT solutions chapter has two exercises and a miscellaneous section. Exercise 8.1 covers the theorem and expansion. Exercise 8.2 focuses on the general term, middle term, and applications. The miscellaneous exercise combines these and adds problems involving coefficients, greatest term, and substitution tricks. These solutions are written to show the general term method for every applicable problem — because it is faster, cleaner, and directly what CBSE mark schemes reward.
Download PDF – All Exercises of NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorey
| Exercise | Topic Covered | Number of Questions |
|---|---|---|
| Exercise 8.1 | Statement of Theorem; Expanding Binomials | 14 Questions |
| Exercise 8.2 | General Term, Middle Term, Applications | 12 Questions |
| Miscellaneous | Coefficient Problems, Advanced Applications | 10 Questions |
Chapter 8 – Binomial Theorem: Concepts, Explanation and Key Tables
The Binomial Theorem Statement
For any positive integer n and real numbers a and b:
(a + b)ⁿ = Σ ⁿCᵣ × aⁿ⁻ʳ × bʳ (where r goes from 0 to n)
Written out, the expansion is: (a + b)ⁿ = ⁿC₀ aⁿb⁰ + ⁿC₁ aⁿ⁻¹b¹ + ⁿC₂ aⁿ⁻²b² + ... + ⁿCₙ a⁰bⁿ
The coefficients ⁿC₀, ⁿC₁, ..., ⁿCₙ form the nth row of Pascal's Triangle.
Properties of Binomial Expansion
| Property | Statement |
|---|---|
| Number of terms | Always (n + 1) terms in the expansion of (a + b)ⁿ |
| Sum of coefficients | Put a = b = 1: sum = 2ⁿ |
| Sum of odd-positioned – even-positioned coefficients | Put a = 1, b = –1: result = 0, so sum of odd = sum of even = 2ⁿ⁻¹ |
| Index of a decreases | Power of a decreases from n to 0 across terms |
| Index of b increases | Power of b increases from 0 to n across terms |
| Symmetry of coefficients | ⁿCᵣ = ⁿCₙ₋ᵣ, so first and last coefficients are equal, second and second-last are equal, etc. |
The General Term — The Most Important Formula in This Chapter
The (r + 1)th term (denoted Tᵣ₊₁) of the expansion (a + b)ⁿ is:
Tᵣ₊₁ = ⁿCᵣ × aⁿ⁻ʳ × bʳ
This single formula solves almost every problem in Exercise 8.2. To find a specific term, set r to the appropriate value. To find a term with a specific power, set the power of x equal to the desired value and solve for r.
Finding the Middle Term
| Case | Number of Terms | Middle Term(s) |
|---|---|---|
| n is even | (n + 1) is odd | Single middle term: T(n/2 + 1) |
| n is odd | (n + 1) is even | Two middle terms: T((n+1)/2) and T((n+3)/2) |
Pascal's Triangle — First 7 Rows
| n | Coefficients (ⁿC₀, ⁿC₁, ..., ⁿCₙ) |
|---|---|
| 0 | 1 |
| 1 | 1 1 |
| 2 | 1 2 1 |
| 3 | 1 3 3 1 |
| 4 | 1 4 6 4 1 |
| 5 | 1 5 10 10 5 1 |
| 6 | 1 6 15 20 15 6 1 |
Each number is the sum of the two numbers directly above it. The outer edges are always 1.
Special Expansions to Know
| Expansion | First Four Terms | Note |
|---|---|---|
| (1 + x)ⁿ | 1 + nx + n(n–1)x²/2! + n(n–1)(n–2)x³/3! + ... | Set a=1, b=x in general formula |
| (1 – x)ⁿ | 1 – nx + n(n–1)x²/2! – n(n–1)(n–2)x³/3! + ... | Alternate signs |
| (a – b)ⁿ | Same as (a+b)ⁿ but odd-powered terms of b are negative | Put –b in place of b |
| (x + 1/x)ⁿ | Terms involve xⁿ, xⁿ⁻², xⁿ⁻⁴ ... descending | Common in coefficient problems |
Applications of the General Term — Worked Strategy
For problems asking "find the coefficient of x⁵ in (2x – 3)⁸":
- Write the general term: Tᵣ₊₁ = ⁸Cᵣ × (2x)⁸⁻ʳ × (–3)ʳ
- Separate powers: = ⁸Cᵣ × 2⁸⁻ʳ × (–3)ʳ × x⁸⁻ʳ
- For x⁵: set 8 – r = 5, so r = 3
- Substitute r = 3: T₄ = ⁸C₃ × 2⁵ × (–3)³ = 56 × 32 × (–27) = –48384
- Coefficient of x⁵ is –48384
Study Tips for Chapter 8
- Memorise the general term formula Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ before attempting any exercise. Every problem in Exercise 8.2 and the miscellaneous exercise uses it.
- When finding the middle term of an expansion where n is odd, both middle terms are worth finding — CBSE sometimes asks for both.
- For binomials with coefficients like (2x + 3)ⁿ, write the general term carefully keeping the coefficients (2 and 3) inside their powers — dropping them is the most common and costly error in this chapter.
- The sum of coefficients shortcut (substitute x = 1) is extremely fast for multiple-choice style questions asking for the sum of all coefficients.
- Practise the "greatest term" application problems from the miscellaneous exercise — these require setting Tᵣ₊₁/Tᵣ ≥ 1 and solving for r, which is a unique technique not seen elsewhere.