## RD Sharma Solutions Class 11 Maths Chapter 21 â€“ Get Free PDF Updated for 2021-22

**RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series**Â are provided here for students to score good marks in the board exams. In this chapter, we intend to discuss the sum to â€˜nâ€™ terms of some other special series viz. series of natural numbers, series of square of natural numbers, series of cubes of natural numbers, etc. **RD Sharma Solutions** are basically designed for CBSE students, in accordance with the latest syllabus prescribed by the CBSE Board.

Our expert team has designed these solutions in an easily understandable language. The practice is an essential task to learn and score well in Mathematics. The pdfs of RD Sharma Class 11 Maths Solutions are provided here. Students are required to go through **RD Sharma Solutions** thoroughly before the final exams to score well and intensify their problem-solving abilities.

**Chapter 21 – Some Special Series** contains two exercises and the RD Sharma Solutions present in this page provide solutions to the questions present in each exercise. Now, let us have a look at the concepts discussed in this chapter.

- Sum to â€˜nâ€™ terms of some special series.
- Sum of the first â€˜nâ€™ natural numbers.
- Sum of the squares of first â€˜nâ€™ natural numbers.
- Sum of the cubes of first â€˜nâ€™ natural numbers.

- Method of difference.
- Summation of some special series.

## Download the pdf of RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series

### Access answers to RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series

EXERCISE 21.1 PAGE NO: 21.10

**Find the sum of the following series to n terms: **

1. 1^{3}Â + 3^{3}Â + 5^{3}Â + 7^{3}Â + â€¦â€¦..

**Solution:**

LetÂ T_{n}Â be theÂ nth term of the given series.

We have:

T_{n} = [1 + (n – 1)2]^{3}

= (2n – 1)^{3}

= (2n)^{3} â€“ 3 (2n)^{2}. 1 + 3.1^{2}.2n-1^{3 }[Since, (a-b)^{3} = a^{3} â€“ 3a^{2}b + 3ab^{2} – b]

= 8n^{3} â€“ 12n^{2} + 6n â€“ 1

Now, letÂ S_{n}Â be the sum ofÂ nÂ terms of the given series.

We have:

Upon simplification we get,

= 2n^{2} (n + 1)^{2} â€“ n â€“ 2n (n + 1) (2n + 1) + 3n (n + 1)

= n (n + 1) [2n (n + 1) â€“ 2 (2n + 1) + 3] â€“ n

= n (n + 1) [2n^{2} â€“ 2n + 1] â€“ n

= n [2n^{3} â€“ 2n^{2} + n + 2n^{2} â€“ 2n + 1 – 1]

= n [2n^{3} – n]

= n^{2} [2n^{2} – 1]

âˆ´ The sum of the series is n^{2} [2n^{2} – 1]

**2. 2 ^{3}Â + 4^{3}Â + 6^{3}Â + 8^{3}Â + â€¦â€¦â€¦**

**Solution:**

LetÂ T_{n}Â be theÂ nth term of the given series.

We have:

T_{n} = (2n)^{3}

= 8n^{3}

Now, letÂ S_{n}Â be the sum ofÂ nÂ terms of the given series.

We have:

âˆ´ The sum of the series is 2{n (n + 1)}^{2}

**3. 1.2.5 + 2.3.6 + 3.4.7 + â€¦â€¦..**

**Solution:**

LetÂ T_{n}Â be theÂ nth term of the given series.

We have:

T_{n} = n (n + 1) (n + 4)

= n (n^{2} + 5n + 4)

= n^{3} + 5n^{2} + 4n

Now, letÂ S_{n}Â be the sum ofÂ nÂ terms of the given series.

We have:

**4. 1.2.4 + 2.3.7 + 3.4.10 + â€¦ to n terms.**

**Solution:**

LetÂ T_{n}Â be theÂ nth term of the given series.

We have:

T_{n} = n (n + 1) (3n + 1)

= n (3n^{2} + 4n + 1)

= 3n^{3} + 4n^{2} + n

Now, letÂ S_{n}Â be the sum ofÂ nÂ terms of the given series.

We have:

âˆ´ The sum of the series is

**5. 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + â€¦ to n terms**

**Solution:**

LetÂ T_{n}Â be theÂ nth term of the given series.

We have:

T_{n} = n(n+1)/2

= (n^{2} + n)/2

Now, letÂ S_{n}Â be the sum ofÂ nÂ terms of the given series.

We have:

âˆ´ The sum of the series is [n(n+1)(n+2)]/6

EXERCISE 21.2 PAGE NO: 21.18

**Sum the following series to n terms:**

**1. 3 + 5 + 9 + 15 + 23 + â€¦â€¦â€¦â€¦.**

**Solution:**

LetÂ T_{n}Â be theÂ nth term and S_{n} be the sum to n terms of the given series.

We have,

S_{n} = 3 + 5 + 9 + 15 + 23 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦ (1)

Equation (1) can be rewritten as:

S_{n} = 3 + 5 + 9 + 15 + 23 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦â€¦..(2)

By subtracting (2) from (1) we get

S_{n} = 3 + 5 + 9 + 15 + 23 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

S_{n} = 3 + 5 + 9 + 15 + 23 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

0 = 3 + [2 + 4 + 6 + 8 + â€¦ + (T_{n} â€“ T_{n-1})] – T_{n}

The difference between the successive terms are 5-3 = 2, 9-5 = 4, 15-9 = 6,

So these differences are in A.P

Now,

âˆ´ The sum of the series is n/3 (n^{2} + 8)

**2. 2 + 5 + 10 + 17 + 26 + â€¦â€¦â€¦..**

**Solution:**

LetÂ T_{n}Â be theÂ nth term and S_{n} be the sum to n terms of the given series.

We have,

S_{n} = 2 + 5 + 10 + 17 + 26 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦ (1)

Equation (1) can be rewritten as:

S_{n} = 2 + 5 + 10 + 17 + 26 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦â€¦..(2)

By subtracting (2) from (1) we get

S_{n} = 2 + 5 + 10 + 17 + 26 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

S_{n} = 2 + 5 + 10 + 17 + 26 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

0 = 2 + [3 + 5 + 7 + 9 + â€¦ + (T_{n} â€“ T_{n-1})] – T_{n}

The difference between the successive terms are 3, 5, 7, 9

So these differences are in A.P

Now,

âˆ´ The sum of the series is n/6 (2n^{2} + 3n + 7)

**3. 1 + 3 + 7 + 13 + 21 + â€¦**

**Solution:**

LetÂ T_{n}Â be theÂ nth term and S_{n} be the sum to n terms of the given series.

We have,

S_{n} = 1 + 3 + 7 + 13 + 21 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦ (1)

Equation (1) can be rewritten as:

S_{n} = 1 + 3 + 7 + 13 + 21 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦â€¦..(2)

By subtracting (2) from (1) we get

S_{n} = 1 + 3 + 7 + 13 + 21 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

S_{n} = 1 + 3 + 7 + 13 + 21 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

0 = 1 + [2 + 4 + 6 + 8 + â€¦ + (T_{n} â€“ T_{n-1})] – T_{n}

The difference between the successive terms are 2, 4, 6, 8

So these differences are in A.P

Now,

âˆ´ The sum of the series is n/3 (n^{2} + 2)

**4. 3 + 7 + 14 + 24 + 37 + â€¦**

**Solution:**

LetÂ T_{n}Â be theÂ nth term and S_{n} be the sum to n terms of the given series.

We have,

S_{n} = 3 + 7 + 14 + 24 + 37 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦ (1)

Equation (1) can be rewritten as:

S_{n} = 3 + 7 + 14 + 24 + 37 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦â€¦..(2)

By subtracting (2) from (1) we get

S_{n} = 3 + 7 + 14 + 24 + 37 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

S_{n} = 3 + 7 + 14 + 24 + 37 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

0 = 3 + [4 + 7 + 10 + 13 + â€¦ + (T_{n} â€“ T_{n-1})] – T_{n}

The difference between the successive terms are 4, 7, 10, 13

So these differences are in A.P

Now,

âˆ´ The sum of the series is n/2 [n^{2} + n + 4]

**5. 1 + 3 + 6 + 10 + 15 + â€¦ **

**Solution:**

LetÂ T_{n}Â be theÂ nth term and S_{n} be the sum to n terms of the given series.

We have,

S_{n} = 1 + 3 + 6 + 10 + 15 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦ (1)

Equation (1) can be rewritten as:

S_{n} = 1 + 3 + 6 + 10 + 15 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n} â€¦â€¦..(2)

By subtracting (2) from (1) we get

S_{n} = 1 + 3 + 6 + 10 + 15 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

S_{n} = 1 + 3 + 6 + 10 + 15 + â€¦â€¦â€¦â€¦. + T_{n-1} + T_{n}

0 = 1 + [2 + 3 + 4 + 5 + â€¦ + (T_{n} â€“ T_{n-1})] – T_{n}

The difference between the successive terms are 2, 3, 4, 5

So these differences are in A.P

Now,

âˆ´ The sum of the series is n/6 (n+1) (n+2)

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