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, and 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 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.

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

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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 T_n be the nth term of the given series.

We have:

T_n = [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 S_n 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 T_n be the nth term of the given series.

We have:

T_n = (2n)³

= 8n³

Now, let S_n 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 T_n be the nth term of the given series.

We have:

T_n = n (n + 1) (n + 4)

= n (n² + 5n + 4)

= n³ + 5n² + 4n

Now, let S_n 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 T_n be the nth term of the given series.

We have:

T_n = n (n + 1) (3n + 1)

= n (3n² + 4n + 1)

= 3n³ + 4n² + n

Now, let S_n 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 T_n be the nth term of the given series.

We have:

T_n = n(n+1)/2

= (n² + n)/2

Now, let S_n be the sum of n terms of the given series.

We have:

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