 # NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.4

NCERT Solutions for Class 11 Maths Chapter 9 have been carefully compiled and developed as per the latest update on the term – I CBSE Syllabus 2021-22. These solutions contain detailed step-by-step explanation of all the problems that are present in the Class 11 NCERT Textbook. Exercise 9.4 of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series is based on the topic Sum to n Terms of Special Series. In this exercise, different types of series are given whose sum to n terms has to be found out.

The NCERT Solutions for Class 11 Maths enhance topics with engaging Math activities that strengthen the concepts further. Each question of the exercise has been carefully solved for the students to understand, keeping the term – I examination point of view in mind.

### Download PDF of NCERT Solutions for Class 11 Maths Chapter 9- Sequences and Series Exercise 9.4       ### Access other exercise solutions of Class 11 Maths Chapter 9- Sequences and Series

Exercise 9.1 Solutions 14 Questions

Exercise 9.2 Solutions 18 Questions

Exercise 9.3 Solutions 32 Questions

Exercise 9.4 Solutions 10 Questions

Miscellaneous Exercise On Chapter 9 Solutions 32 Questions

#### Access Solutions for Class 11 Maths Chapter 9.4 Exercise

Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Solution:

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1)

Then, the sum of n terms of the series can be expressed as 2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Solution:

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

nth term, an = n ( n + 1) ( n + 2)

= (n2 + n) (n + 2)

= n+ 3n+ 2n

Then, the sum of n terms of the series can be expressed as 3. 3 × 12 + 5 × 22 + 7 × 32 + …

Solution:

Given series is 3 ×12 + 5 × 22 + 7 × 32 + …

It’s seen that,

nth term, an = ( 2n + 1) n2 = 2n3 + n2

Then, the sum of n terms of the series can be expressed as 4. Find the sum to n terms of the series Solution: 5. Find the sum to n terms of the series 52 + 62 + 72 + … + 202

Solution:

Given series is 52 + 62 + 72 + … + 202

It’s seen that,

nth term, an = ( n + 4)2 = n2 + 8n + 16

Then, the sum of n terms of the series can be expressed as 6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Solution:

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

a= (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n2 + 15n

Then, the sum of n terms of the series can be expressed as 7. Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …

Solution:

Given series is 12 + (12 + 22) + (12 + 2+ 32 ) + …

Finding the nth term, we have

an = (12 + 22 + 32 +…….+ n2) Now, the sum of n terms of the series can be expressed as 8. Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).

Solution:

Given,

an = n (n + 1) (n + 4) = n(n+ 5n + 4) = n3 + 5n2 + 4n

Now, the sum of n terms of the series can be expressed as 9. Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Solution:

Given,

nth term of the series as:

an = n2 + 2n

Then, the sum of n terms of the series can be expressed as 10. Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Solution:

Given,

nth term of the series as:

an = (2n – 1)2 = 4n2 – 4n + 1

Then, the sum of n terms of the series can be expressed as 