RD Sharma Solutions for Class 11 Mathematics Chapter “Binomial Theorem”

RD Sharma Solutions for Class 11 Mathematics Chapter 18 “Binomial Theorem” are available here, prepared by expert teachers according to the latest CBSE Class 11 Maths Syllabus and NCERT guidelines. This chapter introduces the fundamental concept of the binomial theorem, which is essential for expanding algebraic expressions raised to positive integral powers. On this page, students will find a variety of binomial theorem problems with detailed, step-by-step solutions to help master this important topic.

These RD Sharma Dolution for Class 11 Binomial Theorem solutions will assist students in learning how to expand expressions like (a+b)n, find the general term, and calculate specific coefficients in the expansion. Students will also understand properties of binomial coefficients and the use of Pascal’s Triangle. Mastering the binomial theorem is crucial for excelling in board exams, competitive exams, and further studies in mathematics and related fields.

RD Sharma Solutions for Class 11 Chapter 18 Binomial Theorem

In RD Sharma Solutions for Class 11 Chapter Binomial Theorem, students will gain a clear understanding of how to expand binomial expressions, identify the general term, and apply properties of binomial coefficients. The chapter covers the binomial theorem for positive integral indices, the general term formula, middle term(s) in binomial expansions, and important properties such as symmetry and sum of coefficients. Students also learn to solve problems involving the identification of specific terms and coefficients in binomial expansions, as well as applications in real-life and higher mathematics.

Important topics include the expansion of (a+b)n, the use of binomial coefficients , finding the general and middle terms, and solving a variety of problems based on the binomial theorem. These step-by-step solutions to all textbook exercises provide thorough practice and help build a strong foundation in algebraic expansions, enabling students to confidently solve complex problems in both board exams and competitive tests.

Download RD Sharma Solutions for Binomial Theorem PDF Free

Download comprehensive RD Sharma solutions for the chapter on the binomial theorem, including solved examples, extra practice questions, and detailed explanations. Access the free PDF to enhance your understanding of the binomial theorem, improve accuracy, and boost your exam preparation with trusted, exam-oriented solutions.

Question 1: Expand the binomial expression (a + b)⁵ using the binomial theorem.

Solution:

Step 1: Apply the binomial theorem: (a + b)ⁿ = Σ C(n,r) · a^n-r · b^r for r = 0 to n

Step 2: For n = 5, we have:

(a + b)⁵ = C(5,0)a⁵b⁰ + C(5,1)a⁴b¹ + C(5,2)a³b² + C(5,3)a²b³ + C(5,4)a¹b⁴ + C(5,5)a⁰b⁵

Step 3: Calculate binomial coefficients:

• C(5,0) = 1
• C(5,1) = 5
• C(5,2) = 10
• C(5,3) = 10
• C(5,4) = 5
• C(5,5) = 1

Answer: (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵

Question 2: Find the general term in the expansion of (x + y)⁸.

Solution:

Step 1: The general term in the expansion of (x + y)ⁿ is given by:

T_r+1 = C(n,r) · x^n-r · y^r

Step 2: For (x + y)⁸, n = 8

Answer: T_r+1 = C(8,r) · x^8-r · y^r, where r = 0, 1, 2, ..., 8

Question 3: Calculate the middle term in the expansion of (2x - 3)⁶.

Solution:

Step 1: For (a + b)ⁿ where n is even, there is one middle term at position (n/2 + 1)

Step 2: Here n = 6 (even), so middle term is T₄ (r = 3)

Step 3: T₄ = C(6,3) · (2x)^6-3 · (-3)³

Step 4: Calculate:

• C(6,3) = 20
• (2x)³ = 8x³
• (-3)³ = -27

Answer: Middle term = 20 × 8x³ × (-27) = -4320x³

Question 4: Determine the coefficient of x³ in the expansion of (1 + 2x)⁷.

Solution:

Step 1: General term: T_r+1 = C(7,r) · 1^7-r · (2x)^r

Step 2: T_r+1 = C(7,r) · (2x)^r = C(7,r) · 2^r · x^r

Step 3: For coefficient of x³, we need r = 3

Step 4: T₄ = C(7,3) · 2³ · x³

Step 5: Calculate: C(7,3) = 35, 2³ = 8

Answer: Coefficient of x³ = 35 × 8 = 280

Question 5: Prove the symmetry property of binomial coefficients: C(n,r) = C(n,n-r).

Solution:

Step 1: Start with the definition of binomial coefficient:

C(n,r) = n!/(r!(n-r)!)

Step 2: Calculate C(n,n-r):

C(n,n-r) = n!/((n-r)!(n-(n-r))!) = n!/((n-r)!r!)

Step 3: Since multiplication is commutative:

n!/(r!(n-r)!) = n!/((n-r)!r!)

Answer: Therefore, C(n,r) = C(n,n-r) ✓

Question 6: Find the sum of all the binomial coefficients in the expansion of (1 + x)¹⁰.

Solution:

Step 1: The expansion is: (1 + x)¹⁰ = C(10,0) + C(10,1)x + C(10,2)x² + ... + C(10,10)x¹⁰

Step 2: To find sum of coefficients, substitute x = 1:

(1 + 1)¹⁰ = C(10,0) + C(10,1) + C(10,2) + ... + C(10,10)

Step 3: Calculate: (1 + 1)¹⁰ = 2¹⁰

Answer: Sum of all binomial coefficients = 2¹⁰ = 1024

Question 7: Use Pascal's triangle to find the coefficients in the expansion of (x + y)⁴.

Solution:

Step 1: Pascal's triangle for n = 4:

        1       1 1      1 2 1     1 3 3 1    1 4 6 4 1    

Step 2: The coefficients for (x + y)⁴ are from row 4: 1, 4, 6, 4, 1

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

Question 8: Find the term independent of x in the expansion of (x + 1/x)⁶.

Solution:

Step 1: General term: T_r+1 = C(6,r) · x^6-r · (1/x)^r

Step 2: Simplify: T_r+1 = C(6,r) · x^6-r · x^-r = C(6,r) · x^6-2r

Step 3: For term independent of x, the power of x must be 0:

6 - 2r = 0

r = 3

Step 4: T₄ = C(6,3) = 20

Answer: Term independent of x = 20

Question 9: Calculate the coefficient of x⁴y³ in the expansion of (2x + 3y)⁷.

Solution:

Step 1: General term: T_r+1 = C(7,r) · (2x)^7-r · (3y)^r

Step 2: For x⁴y³, we need:

• Power of x: 7-r = 4, so r = 3
• Power of y: r = 3 ✓

Step 3: T₄ = C(7,3) · (2x)⁴ · (3y)³

Step 4: Calculate:

• C(7,3) = 35
• (2x)⁴ = 16x⁴
• (3y)³ = 27y³

Answer: Coefficient = 35 × 16 × 27 = 15120

Question 10: Explain the binomial theorem for positive integral indices.

Solution:

Statement: For any positive integer n and any real numbers a and b:

(a + b)ⁿ = Σ_r=0ⁿ C(n,r) · a^n-r · b^r

Expanded form:

(a + b)ⁿ = C(n,0)aⁿ + C(n,1)a^n-1b + C(n,2)a^n-2b² + ... + C(n,n)bⁿ

Where:

• C(n,r) = n!/(r!(n-r)!) is the binomial coefficient
• The expansion has (n+1) terms
• The sum of powers in each term equals n

Question 11: Find the coefficient of the middle term in the expansion of (3 + 2x)⁸.

Solution:

Step 1: For n = 8 (even), middle term is T₅ (r = 4)

Step 2: T₅ = C(8,4) · 3^8-4 · (2x)⁴

Step 3: Calculate:

• C(8,4) = 70
• 3⁴ = 81
• (2x)⁴ = 16x⁴

Answer: Coefficient = 70 × 81 × 16 = 90720

Question 12: Determine the number of terms in the expansion of (a + b)ⁿ.

Solution:

Step 1: In the expansion (a + b)ⁿ, terms are:

T₁, T₂, T₃, ..., T_n+1

Step 2: The general term is T_r+1 where r goes from 0 to n

Step 3: Values of r: 0, 1, 2, 3, ..., n (total of n+1 values)

Answer: Number of terms = n + 1

Question 13: Find the coefficient of x⁵ in the expansion of (1 - x)¹⁰.

Solution:

Step 1: General term: T_r+1 = C(10,r) · 1^10-r · (-x)^r

Step 2: T_r+1 = C(10,r) · (-1)^r · x^r

Step 3: For coefficient of x⁵, r = 5

Step 4: T₆ = C(10,5) · (-1)⁵ · x⁵

Step 5: Calculate: C(10,5) = 252, (-1)⁵ = -1

Answer: Coefficient of x⁵ = 252 × (-1) = -252

Question 14: Calculate the sum of the coefficients in the expansion of (2 + 3)ⁿ.

Solution:

Step 1: The expansion of (2 + 3)ⁿ = 5ⁿ

Step 2: But if we consider (2 + 3x)ⁿ and want sum of coefficients:

Step 3: Sum of coefficients is found by substituting x = 1

Step 4: (2 + 3×1)ⁿ = (2 + 3)ⁿ = 5ⁿ

Answer: Sum of coefficients = 5ⁿ

Question 15: Find the general term and the 5th term in the expansion of (x + 2)⁶.

Solution:

Step 1: General term: T_r+1 = C(6,r) · x^6-r · 2^r

Step 2: For the 5th term, r = 4 (since T₅ = T₄₊₁)

Step 3: T₅ = C(6,4) · x^6-4 · 2⁴

Step 4: Calculate:

• C(6,4) = 15
• x²
• 2⁴ = 16

Answer:

• General term: T_r+1 = C(6,r) · x^6-r · 2^r
• 5th term: T₅ = 240x²

Question 16: Explain how to find the middle term when n is odd in a binomial expansion.

Solution:

When n is odd:

Step 1: There are (n+1) terms in the expansion, which is even

Step 2: There are two middle terms:

• T_(n+1)/2 and T_(n+3)/2
• Or equivalently: T_((n+1)/2) and T_((n+1)/2)+1

Step 3: These correspond to r values of (n-1)/2 and (n+1)/2

Example: For n = 5, middle terms are T₃ and T₄ (r = 2 and r = 3)

Question 17: Find the coefficient of x²y³ in the expansion of (x + y)⁵.

Solution:

Step 1: General term: T_r+1 = C(5,r) · x^5-r · y^r

Step 2: For x²y³, we need:

• Power of x: 5-r = 2, so r = 3
• Power of y: r = 3 ✓

Step 3: T₄ = C(5,3) · x² · y³

Step 4: Calculate: C(5,3) = 10

Answer: Coefficient of x²y³ = 10

Question 18: Use the binomial theorem to approximate (1 + x)ⁿ for small values of x.

Solution:

Step 1: Expand (1 + x)ⁿ using binomial theorem:

(1 + x)ⁿ = 1 + nx + (n(n-1)/2!)x² + (n(n-1)(n-2)/3!)x³ + ...

Step 2: For small values of x, higher powers of x become negligible

Step 3: First-order approximation: (1 + x)ⁿ ≈ 1 + nx

Step 4: Second-order approximation: (1 + x)ⁿ ≈ 1 + nx + (n(n-1)/2)x²

Examples:

• √(1 + x) = (1 + x)^1/2 ≈ 1 + x/2
• (1 + x)^-1 ≈ 1 - x

Question 19: Find the coefficient of x³ in the expansion of (1 + 3x)⁵.

Solution:

Step 1: General term: T_r+1 = C(5,r) · 1^5-r · (3x)^r

Step 2: T_r+1 = C(5,r) · (3x)^r = C(5,r) · 3^r · x^r

Step 3: For coefficient of x³, r = 3

Step 4: T₄ = C(5,3) · 3³ · x³

Step 5: Calculate: C(5,3) = 10, 3³ = 27

Answer: Coefficient of x³ = 10 × 27 = 270

Question 20: Prove that the sum of the coefficients in the expansion of (1 + x)ⁿ is 2ⁿ.

Solution:

Step 1: The expansion of (1 + x)ⁿ is:

(1 + x)ⁿ = C(n,0) + C(n,1)x + C(n,2)x² + ... + C(n,n)xⁿ

Step 2: The coefficients are: C(n,0), C(n,1), C(n,2), ..., C(n,n)

Step 3: To find sum of coefficients, substitute x = 1:

(1 + 1)ⁿ = C(n,0) + C(n,1) + C(n,2) + ... + C(n,n)

Step 4: Simplify left side: (1 + 1)ⁿ = 2ⁿ

Answer: Therefore, sum of coefficients = C(n,0) + C(n,1) + ... + C(n,n) = 2ⁿ