RD Sharma Solutions Class 11 Maths Chapter 21 – Get Free PDF Updated for 2023-24
RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series are provided here for students to understand the concepts in detail and score good marks in the board exams. In this chapter, we intend to discuss the sum to ‘n’ terms of some other special series, viz., series of natural numbers, series of the 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.
Chapter 21 – Some Special Series contains two exercises, and the RD Sharma Solutions help students with descriptive answers to the questions present in each exercise. Our expert team has designed these solutions in an easily understandable language. Practice is an essential task to learn and score well in Mathematics. The PDFs of RD Sharma Class 11 Maths Solutions are provided here with the purpose of helping students score well and intensify their problem-solving abilities. Now, let us have a look at the concepts discussed in this chapter.
- Sum to ‘n’ terms of some special series.
- Sum of the first ‘n’ natural numbers.
- Sum of the squares of first ‘n’ natural numbers.
- Sum of the cubes of first ‘n’ natural numbers.
- Method of difference.
- Summation of some special series.
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. 13 + 33 + 53 + 73 + ……..
Solution:
Let Tn be the nth term of the given series.
We have:
Tn = [1 + (n – 1)2]3
= (2n – 1)3
= (2n)3 – 3 (2n)2. 1 + 3.12.2n-13 [Since, (a-b)3 = a3 – 3a2b + 3ab2 – b]
= 8n3 – 12n2 + 6n – 1
Now, let Sn be the sum of n terms of the given series.
We have:
Upon simplification, we get,
= 2n2 (n + 1)2 – n – 2n (n + 1) (2n + 1) + 3n (n + 1)
= n (n + 1) [2n (n + 1) – 2 (2n + 1) + 3] – n
= n (n + 1) [2n2 – 2n + 1] – n
= n [2n3 – 2n2 + n + 2n2 – 2n + 1 – 1]
= n [2n3 – n]
= n2 [2n2 – 1]
∴ The sum of the series is n2 [2n2 – 1]
2. 23 + 43 + 63 + 83 + ………
Solution:
Let Tn be the nth term of the given series.
We have:
Tn = (2n)3
= 8n3
Now, let Sn be the sum of n terms of the given series.
We have:
∴ The sum of the series is 2{n (n + 1)}2
3. 1.2.5 + 2.3.6 + 3.4.7 + ……..
Solution:
Let Tn be the nth term of the given series.
We have:
Tn = n (n + 1) (n + 4)
= n (n2 + 5n + 4)
= n3 + 5n2 + 4n
Now, let Sn be the sum of n terms of the given series.
We have:
4. 1.2.4 + 2.3.7 + 3.4.10 + … to n terms.
Solution:
Let Tn be the nth term of the given series.
We have:
Tn = n (n + 1) (3n + 1)
= n (3n2 + 4n + 1)
= 3n3 + 4n2 + n
Now, let Sn be the sum of n terms of the given series.
We have:
∴ The sum of the series is
5. 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + … to n terms
Solution:
Let Tn be the nth term of the given series.
We have:
Tn = n(n+1)/2
= (n2 + n)/2
Now, let Sn be the sum of n terms of the given series.
We have:
∴ 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 + ………….
Solution:
Let Tn be the nth term, and Sn be the sum to n terms of the given series.
We have,
Sn = 3 + 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 3 + 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1) we get
Sn = 3 + 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
Sn = 3 + 5 + 9 + 15 + 23 + …………. + Tn-1 + Tn
0 = 3 + [2 + 4 + 6 + 8 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 5-3 = 2, 9-5 = 4, 15-9 = 6,
So these differences are in A.P
Now,
∴ The sum of the series is n/3 (n2 + 8)
2. 2 + 5 + 10 + 17 + 26 + ………..
Solution:
Let Tn be the nth term, and Sn be the sum to n terms of the given series.
We have,
Sn = 2 + 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 2 + 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1) we get
Sn = 2 + 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn
Sn = 2 + 5 + 10 + 17 + 26 + …………. + Tn-1 + Tn
0 = 2 + [3 + 5 + 7 + 9 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 3, 5, 7, 9
So these differences are in A.P
Now,
∴ The sum of the series is n/6 (2n2 + 3n + 7)
3. 1 + 3 + 7 + 13 + 21 + …
Solution:
Let Tn be the nth term, and Sn be the sum to n terms of the given series.
We have,
Sn = 1 + 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 1 + 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1), we get
Sn = 1 + 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn
Sn = 1 + 3 + 7 + 13 + 21 + …………. + Tn-1 + Tn
0 = 1 + [2 + 4 + 6 + 8 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 2, 4, 6, 8
So these differences are in A.P
Now,
∴ The sum of the series is n/3 (n2 + 2)
4. 3 + 7 + 14 + 24 + 37 + …
Solution:
Let Tn be the nth term, and Sn be the sum to n terms of the given series.
We have,
Sn = 3 + 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 3 + 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1), we get
Sn = 3 + 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn
Sn = 3 + 7 + 14 + 24 + 37 + …………. + Tn-1 + Tn
0 = 3 + [4 + 7 + 10 + 13 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 4, 7, 10, 13
So these differences are in A.P
Now,
∴ The sum of the series is n/2 [n2 + n + 4]
5. 1 + 3 + 6 + 10 + 15 + …
Solution:
Let Tn be the nth term, and Sn be the sum to n terms of the given series.
We have,
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1), we get
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
0 = 1 + [2 + 3 + 4 + 5 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 2, 3, 4, 5
So these differences are in A.P
Now,
∴ The sum of the series is n/6 (n+1) (n+2)
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