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

Find the sum of the following series to n terms.(Question 1-5)

Question 1. 1³ + 3³ + 5³ + 7³ + ……

Solution:

Let us assume T_n to be the nth term of the given series.

Now, 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

Also, let us assume S_n to be the sum of n terms of the given series.

Simplifying the equation 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]

Therefore, 

The sum of the series is n² [2n² – 1].

Question 2. 2³ + 4³ + 6³ + 8³ + ………

Solution:

Let us assume T_n to be the nth term of the given series.

Now, we have:

T_n = (2n)³

= 8n³

Also, let us assume S_n to be the sum of n terms of the given series.

Therefore, 

The sum of the given series is 2{n (n + 1)}²

Question 3. 1.2.5 + 2.3.6 + 3.4.7 + ……

Solution:

Let us assume T_n to be the nth term of the given series.

Now, we have:

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

= n (n² + 5n + 4)

= n³ + 5n² + 4n

Also, let us assume S_n to be the sum of n terms of the given series.

Therefore, 

The sum of the given series is 

Question 4. 1.2.4 + 2.3.7 + 3.4.10 + ….

Solution:

Let us assume T_n to be the nth term of the given series.

Now, we have:

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

= n (3n² + 4n + 1)

= 3n³ + 4n² + n

Also, let us assume S_n to be the sum of n terms of the given series.

Therefore, 

The sum of the given series is 

Question 5. 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + …

Solution:

Let us assume T_n to be the nth term of the given series.

Now, we have:

T_n = n(n + 1)/2

= (n² + n)/2

Also, let us assume S_n to be the sum of n terms of the given series.

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

Question 6. Find the sum of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + … upto n terms.

Solution:

Let us assume T_n to be the nth term of the given series.
T_n = (nth term of 1, 2, 3..) x (nth term of 2, 3, 4…)

= [1 + (n + 1) x 1].[2 + (n + 1) x 1]

= [1 + n – 1].[2 + n – 1]

= n(n + 1)

= n² + n 

Also, let us assume S_n to be the sum of n terms of the given series.

Question 7. Find the sum of the series 3 × 1² + 5 × 2² + 7 × 3² + … upto n terms.

Solution:

Let us assume T_n to be the nth term of the given series.

T_n = (nth term of 3, 5, 7..) x (nth term of 1², 2², 3²…)

= [3 + (n – 1) x 2].[n²]

= [2n + 1]. [n²]

= 2n³ + n²

T_n = 2n³ + n²

Also, let us assume S_n to be the sum of n terms of the given series.

Therefore, 

The sum of the series = 

Question 8 (i). Find the sum of the series 2n³ + 3n² – 1 to n terms.

Solution:

T_n = 2n³ + 3n² – 1 

Also, let us assume S_n to be the sum of n terms of the given series.

Sum of n terms = 

Question 8 (ii). Find the sum of the series n³ – 3ⁿ to n terms.

Solution:

T_n = n³ – 3ⁿ

Also, let us assume S_n to be the sum of n terms of the given series.

Question 8 (iii). Find the sum of the series n(n + 1)(n + 4) to n terms.

Solution:

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

Also, let us assume S_n to be the sum of n terms of the given series.

Question 8 (iv). Find the sum of the series (2n – 1)² to n terms.

Solution:

T_n = (2n – 1)² 

Also, let us assume S_n to be the sum of n terms of the given series.

Question 9. Find the 20th term and the sum of 20 terms of the series 2 × 4 + 4 × 6 + 6 × 8 +…..  

Solution:

Let us assume T_n to be the nth term of the given series.

The 20th term of the series is :

The infinite series is equivalent to, 

2 × 4 + 4 × 6 + 6 × 8 + …. = 

Sum of the series until 20th term is equivalent to