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Class 11 Maths NCERT Solutions

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Binomial Theorem Class 11 Maths NCERT Solutions

The complete NCERT exercise solutions for Chapter 7, Binomial Theorem — 14 questions from Ex 7.1, each worked through step by step in the CBSE marking pattern. The binomial expansion, the general term, the middle term, and simple applications.

Class:11Subject:MathsChapter:7
3 Key Formulas
DWritten byDeep Narayan
Updated
Key Concept Summary

How many questions are in NCERT Class 11 Maths Chapter 7?

Chapter 7 carries 1 exercise question, numbered Ex 7.1. All of them are solved step by step on this page, along with the chapter's key formulas and exam pointers.

01

Chapter Overview

The binomial theorem expands (x + y)ⁿ as a sum of n+1 terms whose coefficients are the binomial numbers ⁿCᵣ. The expansion is symmetric — the coefficients read the same forwards and backwards (⁰C₀, ⁰C₁, …, ⁰Cₙ = Pascal's row) — and the (r+1)th term is ⁿCᵣ xⁿ⁻ʳ yʳ. That term formula is the workhorse for evaluating numbers like (99)⁵ and for divisibility proofs.

Board pattern

In a written expansion, always pair each term with its coefficient explicitly (write ⁵C₂(1)³(−2x)² before simplifying). For 'indicate which is larger' and approximation questions, the first one or two terms alone often decide the answer — quote the inequality formed by dropping the positive tail.
02

Exercise 7.1 — Using the Binomial Theorem

14Exercise questions

Step-by-step solution

  1. 1(1 − 2x)⁵ = Σ ⁵Cᵣ (1)⁵⁻ʳ (−2x)ʳ.
  2. 2= 1 − 5(2x) + 10(2x)² − 10(2x)³ + 5(2x)⁴ − (2x)⁵.
  3. 3= 1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.

Final answer

1 − 10x + 40x² − 80x³ + 80x⁴ − 32x⁵.

Step-by-step solution

  1. 1(2/x − x/2)⁵ = Σ ⁵Cᵣ (2/x)⁵⁻ʳ (−x/2)ʳ.
  2. 2= 32/x⁵ − 5·16/x⁴·x/2 + 10·8/x³·x²/4 − 10·4/x²·x³/8 + 5·2/x·x⁴/16 − x⁵/32.
  3. 3= 32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

Final answer

32/x⁵ − 40/x³ + 20/x − 5x + 5x³/8 − x⁵/32.

Step-by-step solution

  1. 1(2x − 3)⁶ = Σ ⁶Cᵣ (2x)⁶⁻ʳ (−3)ʳ.
  2. 2= 64x⁶ − 6·32x⁵·3 + 15·16x⁴·9 − 20·8x³·27 + 15·4x²·81 − 6·2x·243 + 729.
  3. 3= 64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

Final answer

64x⁶ − 576x⁵ + 2160x⁴ − 4320x³ + 4860x² − 2916x + 729.

Step-by-step solution

  1. 1(x/3 + 1/x)⁵ = Σ ⁵Cᵣ (x/3)⁵⁻ʳ (1/x)ʳ.
  2. 2= x⁵/243 + 5x⁴/81·1/x + 10x³/27·1/x² + 10x²/9·1/x³ + 5x/3·1/x⁴ + 1/x⁵.
  3. 3= x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

Final answer

x⁵/243 + 5x³/81 + 10x/27 + 10/(9x) + 5/(3x³) + 1/x⁵.

Step-by-step solution

  1. 1(x + 1/x)⁶ = Σ ⁶Cᵣ x⁶⁻ʳ (1/x)ʳ.
  2. 2= x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

Final answer

x⁶ + 6x⁴ + 15x² + 20 + 15/x² + 6/x⁴ + 1/x⁶.

Step-by-step solution

  1. 196 = 100 − 4, so (96)³ = (100 − 4)³.
  2. 2= 100³ − 3·100²·4 + 3·100·4² − 4³.
  3. 3= 1000000 − 120000 + 4800 − 64 = 884736.

Final answer

884736.

Step-by-step solution

  1. 1102 = 100 + 2, so (102)⁵ = (100 + 2)⁵.
  2. 2= 100⁵ + 5·100⁴·2 + 10·100³·4 + 10·100²·8 + 5·100·16 + 32.
  3. 3= 10000000000 + 1000000000 + 40000000 + 800000 + 8000 + 32 = 11040808032.

Final answer

11040808032.

Step-by-step solution

  1. 1101 = 100 + 1, so (101)⁴ = (100 + 1)⁴.
  2. 2= 100⁴ + 4·100³ + 6·100² + 4·100 + 1.
  3. 3= 100000000 + 4000000 + 60000 + 400 + 1 = 104060401.

Final answer

104060401.

Step-by-step solution

  1. 199 = 100 − 1, so (99)⁵ = (100 − 1)⁵.
  2. 2= 100⁵ − 5·100⁴ + 10·100³ − 10·100² + 5·100 − 1.
  3. 3= 10000000000 − 5000000 + 100000 − 10000 + 500 − 1 = 9509900499.

Final answer

9509900499.

Step-by-step solution

  1. 1(1.1)¹⁰⁰⁰⁰ = (1 + 0.1)¹⁰⁰⁰⁰ = 1 + ¹⁰⁰⁰⁰C₁(0.1) + positive terms.
  2. 2= 1 + 10000 × 0.1 + (positive) = 1001 + (positive).
  3. 31001 > 1000, hence (1.1)¹⁰⁰⁰⁰ is larger.

Final answer

(1.1)¹⁰⁰⁰⁰.

Step-by-step solution

  1. 1(a+b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴.
  2. 2(a−b)⁴ = a⁴ − 4a³b + 6a²b² − 4ab³ + b⁴.
  3. 3Difference = 8a³b + 8ab³ = 8ab(a² + b²).
  4. 4With a = √3, b = √2: 8√3√2(3 + 2) = 8√6 × 5 = 40√6.

Final answer

8ab(a² + b²); hence 40√6.

Step-by-step solution

  1. 1Even powers survive the addition: (x+1)⁶ + (x−1)⁶ = 2(x⁶ + 15x⁴ + 15x² + 1).
  2. 2With x = √2: 2[(√2)⁶ + 15(√2)⁴ + 15(√2)² + 1].
  3. 3= 2[8 + 60 + 30 + 1] = 2 × 99 = 198.

Final answer

2(x⁶ + 15x⁴ + 15x² + 1); hence 198.

Step-by-step solution

  1. 19ⁿ⁺¹ = (1 + 8)ⁿ⁺¹ = 1 + (n+1)·8 + ⁰ⁿ⁺¹C₂·8² + … + 8ⁿ⁺¹.
  2. 2= (1 + 8n + 8) + 64·(⁰ⁿ⁺¹C₂ + ⁰ⁿ⁺¹C₃·8 + …) = 9 + 8n + 64k.
  3. 3Then 9ⁿ⁺¹ − 8n − 9 = 64k, which is divisible by 64.

Final answer

9ⁿ⁺¹ − 8n − 9 = 64k (shown).

Step-by-step solution

  1. 1By the binomial theorem, (1 + x)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ xʳ.
  2. 2Put x = 3: (1 + 3)ⁿ = Σᵣ₌₀ⁿ ⁿCᵣ 3ʳ.
  3. 3Hence Σᵣ₌₀ⁿ 3ʳ ⁿCᵣ = 4ⁿ.

Final answer

Proved: LHS = 4ⁿ.

Quick Revision

Key formulas at a glance

Memorise these equations — direct application numericals and derivations in CBSE & JEE frequently hinge on these.

Binomial expansion

General term

Middle term

Exam Strategy

How this chapter is asked

High-yield question patterns observed across CBSE boards, JEE Main & Advanced, and NEET.

  • Every term-sum, coefficient-of-a-particular-power and middle-term question is solved by writing the general term and matching the exponent of x.
  • Expand (1 + x)^n by using x = 1: the sum of all coefficients is 2^n, and the alternating sum is 0 for n positive.

FAQ

Frequently asked questions

How many questions are in NCERT Class 11 Maths Chapter 7 (Binomial Theorem)?

There are 1 exercise question in this chapter, numbered Ex 7.1. Every one is solved step by step on this page in the official NCERT numbering.

Which formulas come up in Binomial Theorem Class 11 Maths?

The formulas this chapter's questions actually turn on are: Binomial expansion, General term, Middle term. They are listed with their expressions in the key formulas section below, and the solved questions show where each one is used.

Is Binomial Theorem important for JEE Main?

Important — general-term and coefficient questions are short and formulaic, worth quick marks in boards and frequently asked in JEE Main.

Next Chapters

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