RD Sharma Solutions for Class 11 Maths Exercise 21.1 Chapter 21 - Some Special Series

In Exercise 21.1 of Chapter 21, we shall discuss problems based on the sum of the first ‘n’ natural numbers, the sum of the squares of first ‘n’ natural numbers, the sum of the cubes of first ‘n’ natural numbers. Students wishing to clear their doubts pertaining to this exercise can utilise the RD Sharma Class 11 Maths Solutions. The solutions are formulated by our expert faculty team in a comprehensive manner to make it interesting for students to solve. The links to the solutions of this exercise can be accessed in the RD Sharma Class 11 Maths pdf, which is available in the below-mentioned links.

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Exercise 21.2 Solutions

Access answers to RD Sharma Solutions for Class 11 Maths Exercise 21.1 Chapter 21 – Some Special Series

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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 1

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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 2

∴ 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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 3

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 4

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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 5

∴ The sum of the series is

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 6

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:

RD Sharma Solutions for Class 11 Maths Chapter 21 – Some Special Series image - 7

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

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