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

• Last Updated : 18 Mar, 2021

### Question 1. 13 + 33 + 53 + 73 + ……

Solution:

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

Now, 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

Also, let us assume Sn 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 Tn to be the nth term of the given series.

Now, we have:

Tn = (2n)3

= 8n3

Also, let us assume Sn 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 Tn to be the nth term of the given series.

Now, we have:

Tn = n (n + 1) (n + 4)

= n (n2 + 5n + 4)

= n3 + 5n2 + 4n

Also, let us assume Sn 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 Tn to be the nth term of the given series.

Now, we have:

Tn = n (n + 1) (3n + 1)

= n (3n2 + 4n + 1)

= 3n3 + 4n2 + n

Also, let us assume Sn 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 Tn to be the nth term of the given series.

Now, we have:

Tn = n(n + 1)/2

= (n2 + n)/2

Also, let us assume Sn 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 Tn to be the nth term of the given series.
Tn = (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 Sn 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 Tn to be the nth term of the given series.

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

= [3 + (n – 1) x 2].[n2]

= [2n + 1]. [n2]

= 2n3 + n2

Tn = 2n3 + n2

Also, let us assume Sn 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:

Tn = 2n3 + 3n2 – 1

Also, let us assume Sn 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:

Tn = n3 – 3n

Also, let us assume Sn 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:

Tn = n(n + 1)(n + 4)

Also, let us assume Sn 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:

Tn = (2n – 1)

Also, let us assume Sn 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 Tn 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

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