RD Sharma Solutions for Class 11 Chapter 21 – Some Special Series

RD Sharma Solutions for Class 11 Chapter 21 – Some Special Series is an essential study resource for students looking to strengthen their understanding of special mathematical series. This chapter builds on the concepts of sequences and series from previous chapters, making it a crucial part of your learning journey in Class 11 mathematics. Special series, like the sum of natural numbers, squares, and cubes, are used in various academic and real-life applications. If you're looking for clear explanations and step-by-step solutions, this chapter is designed to make learning simple and effective.

What Students will Learn in Some Special Series

In this chapter, students explore key special series formulas, including:

• The sum of the first n natural numbers: Sn = n(n+1)/2
• The sum of the squares of the first n natural numbers: Sn = n(n+1)(2n+1)/6
• The sum of the cubes of the first n natural numbers: Sn = [n(n+1)/2]2

Each formula is explained with clear, step-by-step examples. The chapter also shows how to apply these formulas to solve a variety of problems, making it easier to tackle competitive exams and more complex math topics.

Do Check: RD Sharma Solutions for Class 6 to 12

RD Sharma Solutions for Class 11 Chapter 21 provide detailed, step-by-step answers for all textbook problems. Each solution is written in simple, understandable language, with a focus on clarity, so students can easily follow the process. The solutions also include helpful diagrams, solved examples, and tips to boost exam preparation.

By studying these solutions, students not only improve their exam performance but also build a strong foundation in special series concepts. Whether you're preparing for JEE or simply want to improve your understanding, RD Sharma Chapter 21 Solutions are your perfect guide to mastering the topic.

Download RD Sharma Solutions for Class 11 Chapter 21 – Some Special Series PDF 

The RD Sharma Class 11 Chapter 21 PDF contains detailed solutions, solved examples, and extra practice questions to help you better understand special series. 

Chapters to Study Before Some Special Series

Before diving into Chapter 21, it's important to be familiar with the following chapters:

• Chapter 19 – Arithmetic Progressions (A.P.): Learn about sequences where each term increases by a fixed value and how to calculate their sums. Students can check RD Sharma Solutions for Chapter 19 Arithmetic Progression
• Chapter 20 – Geometric Progressions (G.P.): Understand sequences where each term is multiplied by a constant ratio and solve related problems. Check RD Sharma Solutions for Geometric Progressions
• Basic Sequences and Series Concepts: These chapters introduce the foundational language and logic of sequences, which are key to understanding special series.

Access Answers to RD Sharma Solutions for Class 11 Chapter 21 – Some Special Series

This chapter covers special series, their properties, and how to solve related problems. The RD Sharma Solutions for Class 11 are designed to be easy to follow, ensuring that students understand each concept thoroughly and can apply it confidently in exams.

RD Sharma Class 11 Chapter 21 Some Special Series: Exercises

Question 1: Find the sum of the first 50 natural numbers.

Solution:

Formula for sum of first n natural numbers: Sn = n(n+1)/2

Given: n = 50

S50 = 50(50+1)/2

= 50(51)/2

= 2550/2

= 1275

Answer: 1275

Question 2: Calculate the sum of the squares of the first 20 natural numbers.

Solution:

Formula for sum of squares of first n natural numbers: Sn = n(n+1)(2n+1)/6

Given: n = 20

S20 = 20(20+1)(2×20+1)/6

= 20(21)(41)/6

= 17220/6

= 2870

Answer: 2870

Question 3: Find the sum of the cubes of the first 10 natural numbers.

Solution:

Formula for sum of cubes of first n natural numbers: Sn = [n(n+1)/2]²

Given: n = 10

S10 = [10(10+1)/2]²

= [10(11)/2]²

= [110/2]²

= [55]²

= 3025

Answer: 3025

Question 4: If the sum of the first n natural numbers is 210, find the value of n.

Solution:

Given: Sum of first n natural numbers = 210

Using formula: n(n+1)/2 = 210

n(n+1) = 420

n² + n = 420

n² + n - 420 = 0

Factoring: (n + 21)(n - 20) = 0

n = -21 or n = 20

Since n must be positive, n = 20

Answer: n = 20

Question 5: Prove that the sum of the cubes of the first n natural numbers is equal to the square of their sum.

Solution:

We need to prove: 1³ + 2³ + 3³ + ... + n³ = (1 + 2 + 3 + ... + n)²

LHS = Sum of cubes of first n natural numbers = [n(n+1)/2]²

RHS = Square of sum of first n natural numbers = [n(n+1)/2]²

Since LHS = RHS, the statement is proved.

Proof Complete

Question 6: Find the sum: 1² + 2² + 3² + ... + 15².

Solution:

Using formula for sum of squares: Sn = n(n+1)(2n+1)/6

Given: n = 15

S15 = 15(15+1)(2×15+1)/6

= 15(16)(31)/6

= 7440/6

= 1240

Answer: 1240

Question 7: Evaluate the sum: 1³ + 2³ + 3³ + ... + 8³.

Solution:

Using formula for sum of cubes: Sn = [n(n+1)/2]²

Given: n = 8

S8 = [8(8+1)/2]²

= [8(9)/2]²

= [72/2]²

= [36]²

= 1296

Answer: 1296

Question 8: If the sum of the squares of the first n natural numbers is 385, find n.

Solution:

Given: Sum of squares of first n natural numbers = 385

Using formula: n(n+1)(2n+1)/6 = 385

n(n+1)(2n+1) = 2310

Testing values: For n = 10:

10(11)(21) = 2310 ✓

Answer: n = 10

Question 9: Show that the sum of the first n odd natural numbers is n².

Solution:

First n odd natural numbers: 1, 3, 5, ..., (2n-1)

This forms an A.P. with:

First term (a) = 1

Common difference (d) = 2

Number of terms = n

Sum = n/2[2a + (n-1)d]

= n/2[2(1) + (n-1)(2)]

= n/2[2 + 2n - 2]

= n/2(2n)

= n²

Proof Complete

Question 10: Find the sum of the first 25 even natural numbers.

Solution:

First 25 even natural numbers: 2, 4, 6, ..., 50

Formula for sum of first n even natural numbers: Sn = n(n+1)

Given: n = 25

S25 = 25(25+1)

= 25(26)

= 650

Answer: 650

Question 11: Prove that the sum of the first n even natural numbers is n(n + 1).

Solution:

First n even natural numbers: 2, 4, 6, ..., 2n

This forms an A.P. with:

First term (a) = 2

Common difference (d) = 2

Number of terms = n

Sum = n/2[2a + (n-1)d]

= n/2[2(2) + (n-1)(2)]

= n/2[4 + 2n - 2]

= n/2[2n + 2]

= n/2 × 2(n + 1)

= n(n + 1)

Proof Complete

Question 12: Find the sum of the series: 2 + 4 + 6 + ... + 40.

Solution:

This is the sum of first 20 even natural numbers (since 40 = 2 × 20)

Using formula: Sn = n(n+1)

Given: n = 20

S20 = 20(20+1)

= 20(21)

= 420

Answer: 420

Question 13: Calculate the sum of the series: 1 + 3 + 5 + ... + 99.

Solution:

Finding the number of terms: nth odd number = 2n - 1

99 = 2n - 1

2n = 100

n = 50

Sum of first 50 odd numbers = n² = 50² = 2500

Answer: 2500

Question 14: If the sum of the cubes of the first n natural numbers is 44100, find n.

Solution:

Given: Sum of cubes of first n natural numbers = 44100

Using formula: [n(n+1)/2]² = 44100

n(n+1)/2 = √44100 = 210

n(n+1) = 420

n² + n - 420 = 0

(n + 21)(n - 20) = 0

Since n > 0, n = 20

Answer: n = 20

Question 15: Find the value of n if the sum of the first n natural numbers is 630.

Solution:

Given: Sum of first n natural numbers = 630

Using formula: n(n+1)/2 = 630

n(n+1) = 1260

n² + n - 1260 = 0

Using quadratic formula: n = [-1 ± √(1 + 5040)]/2

n = [-1 ± √5041]/2

n = [-1 ± 71]/2

Taking positive value: n = 70/2 = 35

Answer: n = 35

Question 16: Find the sum of the squares of the first 12 even natural numbers.

Solution:

First 12 even natural numbers: 2, 4, 6, ..., 24

Sum = 2² + 4² + 6² + ... + 24²

= 2²(1² + 2² + 3² + ... + 12²)

= 4 × [12(12+1)(2×12+1)/6]

= 4 × [12(13)(25)/6]

= 4 × [3900/6]

= 4 × 650

= 2600

Answer: 2600

Question 17: Prove that 1³ + 2³ + 3³ + ... + n³ = [n(n + 1)/2]².

Solution:

This can be proved by mathematical induction:

Base case (n = 1):

LHS = 1³ = 1

RHS = [1(1+1)/2]² = [1×2/2]² = 1² = 1

LHS = RHS ✓

Inductive step: Assume the formula holds for n = k

1³ + 2³ + ... + k³ = [k(k+1)/2]²

For n = k+1:

1³ + 2³ + ... + k³ + (k+1)³ = [k(k+1)/2]² + (k+1)³

After algebraic manipulation, this equals [(k+1)(k+2)/2]²

Proof Complete

Question 18: Find the sum of the series: 1² + 3² + 5² + ... + (2n – 1)².

Solution:

This is the sum of squares of first n odd numbers.

Formula: Sum = n(2n-1)(2n+1)/3

Derivation: Sum of squares of odd numbers from 1 to (2n-1)

= Sum of all squares from 1 to (2n-1) - Sum of squares of even numbers

After simplification: n(2n-1)(2n+1)/3

Answer: n(2n-1)(2n+1)/3

Question 19: Calculate the sum: 1² + 2² + 3² + ... + n² for n = 30.

Solution:

Using formula for sum of squares: Sn = n(n+1)(2n+1)/6

Given: n = 30

S30 = 30(30+1)(2×30+1)/6

= 30(31)(61)/6

= 56730/6

= 9455

Answer: 9455

Question 20: If the sum of the first n even natural numbers is 420, find n.

Solution:

Given: Sum of first n even natural numbers = 420

Using formula: n(n+1) = 420

n² + n - 420 = 0

Factoring: (n + 21)(n - 20) = 0

n = -21 or n = 20

Since n must be positive, n = 20

Answer: n = 20