Class 11 NCERT Solutions- Chapter 9 Sequences And Series – Exercise 9.4

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

Question 1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Solution:

Given: Series = 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

To find nth term, we have

n^th term, a_n = n ( n + 1)

So, the sum of n terms of the series:

Question 2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Solution:

Given: Series = 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

To find nth term, we have

n^th term, a_n = n(n + 1)(n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

So, the sum of n terms of the series:

Question 3. 3 × 1² + 5 × 2² + 7 × 3² + …

Solution:

Given: Series = 3 × 1² + 5 × 2² + 7 × 3² + …

To find nth term, we have

n^thterm, a_n = (2n + 1) n² = 2n³ + n²

So, the sum of n terms of the series:

Question 4. Find the sum to n terms of the series  

Solution:

Given: Series = 

To find nth term, we have

n^thterm a_n =          (By partial fractions)

So, on adding the above terms columns wise, we obtain

Question 5. Find the sum to n terms of the series 5² + 6² + 7² + … + 20²

Solution:

Given: Series = 5² + 6² + 7² + … + 20²

To find nth term, we have

n^th term, a_n = (n + 4)² = n² + 8n + 16

So, the sum of n terms of the series:

16^th term is (16 + 4) = 20²

= 1496 + 1088 + 256

= 2840

So, 5² + 6² + 7² + …..+ 20² = 2840

Question 6. Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…

Solution:

Given: Series = 3 × 8 + 6 × 11 + 9 × 14 + …

To find nth term, we have

a_n = (n^th term of 3, 6, 9 …) × (n^th term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n² + 15n

So, the sum of n terms of the series:

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

Question 7. Find the sum to n terms of the series 1² + (1² + 2²) + (1² + 2² + 3²) + …

Solution:

Given: Series = 1² + (1² + 2²) + (1² + 2² + 3²⁾ + …

To find nth term, we have

a_n = (1² + 2² + 3² +…….+ n²)

So, the sum of n terms of the series:

Question 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:

Question 9. Find the sum to n terms of the series whose nth terms is given by n² + 2ⁿ

Solution:

Given: n^th term of the series as:

a_n = n² + 2ⁿ

So, the sum of n terms of the series:

Now, the above series 2, 2², 2³ ….. is G.P. 

and the first term and common ration is equal to 2.

From eq(1) and (2), we get

Question 10. Find the sum to n terms of the series whose n^th terms is given by (2n – 1)²

Solution:

Given: n^thterm of the series as:

a_n = (2n – 1)² = 4n² – 4n + 1

So, the sum of n terms of the series: