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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 8.

NCERT Solutions for Class 11 Maths Chapter 9 have been carefully compiled and developed as per the latest update on the CBSE Syllabus 2023-23. These solutions contain detailed step-by-step explanations 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 detailed topics with engaging Math activities that help to strengthen the concepts further. Each question of the exercise has been carefully solved in the NCERT Class 11 Maths Solutions for the students to understand, keeping the board examination point of view in mind.

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

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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

Also explore – NCERT Class 11 Solutions

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,

n^th term, a_n = 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,

n^th term, a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 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,

n^th term, a_n = ( 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,

n^th term, a_n = ( 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_n = (n^th term of 3, 6, 9 …) × (n^th 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 + 2² + 3² ) + …

Finding the n^th term, we have

a_n = (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 n^th term is given by n (n + 1) (n + 4).

Solution:

Given,

a_n = n (n + 1) (n + 4) = n(n² + 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 n^th terms is given by n² + 2ⁿ

Solution:

Given,

n^th term of the series as:

a_n = 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 n^th terms is given by (2n – 1)²

Solution:

Given,

n^th term of the series as:

a_n = (2n – 1)² = 4n² – 4n + 1

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