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

• Last Updated : 21 Feb, 2021

### Question 1. 1 Ã— 2 + 2 Ã— 3 + 3 Ã— 4 + 4 Ã— 5 + â€¦

Solution:

Given: Series = 1 Ã— 2 + 2 Ã— 3 + 3 Ã— 4 + 4 Ã— 5 + â€¦

To find nth term, we have

nth term, an = n ( n + 1)

So, the sum of n terms of the series:

### Question 2. 1 Ã— 2 Ã— 3 + 2 Ã— 3 Ã— 4 + 3 Ã— 4 Ã— 5 + â€¦

Solution:

Given: Series = 1 Ã— 2 Ã— 3 + 2 Ã— 3 Ã— 4 + 3 Ã— 4 Ã— 5 + â€¦

To find nth term, we have

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

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

So, the sum of n terms of the series:

### Question 3. 3 Ã— 12 + 5 Ã— 22 + 7 Ã— 32 + â€¦

Solution:

Given: Series = 3 Ã— 12 + 5 Ã— 22 + 7 Ã— 32 + â€¦

To find nth term, we have

nthterm, an = (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

nthterm an         (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 + â€¦ + 202

Solution:

Given: Series = 52 + 62 + 72 + â€¦ + 202

To find nth term, we have

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

So, the sum of n terms of the series:

16th 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 Ã— 8 + 6 Ã— 11 + 9 Ã— 14 +â€¦

Solution:

Given: Series = 3 Ã— 8 + 6 Ã— 11 + 9 Ã— 14 + â€¦

To find nth term, we have

an = (nth term of 3, 6, 9 â€¦) Ã— (nth term of 8, 11, 14, â€¦)

= (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) + â€¦

Solution:

Given: Series = 12 + (12 + 22) + (12 + 22 + 32) + â€¦

To find nth term, we have

an = (12 + 22 + 32 +â€¦â€¦.+ n2)

So, the sum of n terms of the series:

### Question 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(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: nth term of the series as:

an = 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 nth terms is given by (2n â€“ 1)2

Solution:

Given: nthterm of the series as:

an = (2n â€“ 1)2 = 4n2 â€“ 4n + 1

So, the sum of n terms of the series:

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