NCERT Solutions for Class 11 Maths Chapter 9 – Sequences and Series Ex 9.4
Page No 196:
Question 1:
Find the sum to n terms of the series 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
Answer:
The given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …
nth term, an = n ( n + 1)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4411/chapter%209_html_m4e203d65.gif)
Question 2:
Find the sum to n terms of the series 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
Answer:
The given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …
nth term, an = n ( n + 1) ( n + 2)
= (n2 + n) (n + 2)
= n3 + 3n2 + 2n
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4415/chapter%209_html_7e052a7f.gif)
Question 3:
Find the sum to n terms of the series 3 × 12 + 5 × 22 + 7 × 32 + …
Answer:
The given series is 3 ×12 + 5 × 22 + 7 × 32 + …
nth term, an = ( 2n + 1) n2 = 2n3 + n2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5615/Chapter%209_html_732c903.gif)
Question 3:
Find the sum to n terms of each of the series in Exercises 1 to 7.
3 × 12 + 5 × 22 + 7 × 32 + …
Answer:
The given series is 3 ×12 + 5 × 22 + 7 × 32 + …
nth term, an = ( 2n + 1) n2 = 2n3 + n2
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4419/chapter%209_html_m6a8fa1c7.gif)
Question 4:
Find the sum to n terms of the series ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_4f04d67c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_4f04d67c.gif)
Answer:
The given series is ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_4f04d67c.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_4f04d67c.gif)
nth term, an = ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_2e2c9235.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_403318fc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_2e2c9235.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_403318fc.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_m6bfabef3.gif)
Adding the above terms column wise, we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4421/chapter%209_html_m685ec424.gif)
Question 5:
Find the sum to n terms of the series ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5616/Chapter%209_html_f731ed9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5616/Chapter%209_html_f731ed9.gif)
Answer:
The given series is 52 + 62 + 72 + … + 202
nth term, an = ( n + 4)2 = n2 + 8n + 16
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5616/Chapter%209_html_m7523615f.gif)
16th term is (16 + 4)2 = 2022
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5616/Chapter%209_html_m46d29076.gif)
Question 6:
Find the sum to n terms of the series 3 × 8 + 6 × 11 + 9 × 14 +…
Answer:
The given series is 3 × 8 + 6 × 11 + 9 × 14 + …
an = (nth term of 3, 6, 9 …) × (nth term of 8, 11, 14, …)
= (3n) (3n + 5)
= 9n2 + 15n
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4424/chapter%209_html_3abc8057.gif)
Question 7:
Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + …
Answer:
The given series is 12 + (12 + 22) + (12 + 22 + 32 ) + …
an = (12 + 22 + 32 +…….+ n2)
= nn+12n+16=n2n2+3n+16=2n3+3n2+n6=13n3+12n2+16n
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4425/chapter%209_html_m2e680871.gif)
Question 8:
Find the sum to n terms of the series whose nth term is given by n (n + 1) (n + 4).
Answer:
an = n (n + 1) (n + 4) = n(n2 + 5n + 4) = n3 + 5n2 + 4n
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4426/chapter%209_html_77429ad8.gif)
Question 9:
Find the sum to n terms of the series whose nth terms is given by n2 + 2n
Answer:
an = n2 + 2n
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5617/Chapter%209_html_m2be851a8.gif)
Consider ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5617/Chapter%209_html_42eadf42.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5617/Chapter%209_html_42eadf42.gif)
The above series 2, 22, 23, … is a G.P. with both the first term and common ratio equal to 2.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5617/Chapter%209_html_m9f9f5.gif)
Therefore, from (1) and (2), we obtain
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/5617/Chapter%209_html_m24a0a3c.gif)
Question 10:
Find the sum to n terms of the series whose nth terms is given by (2n – 1)2
Answer:
an = (2n – 1)2 = 4n2 – 4n + 1
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/169/4430/chapter%209_html_m10a972bb.gif)