Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Chapter 9 Class 11 Sequences and Series
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Example 9 Important
Example 10 Important
Ex 8.2, 3 Important
Ex 8.2, 11 Important
Ex 8.2, 17 Important
Ex 8.2, 18 Important
Ex 8.2, 22 Important
Ex 8.2, 28
Ex 8.2, 29 Important
Ex 9.4.4 Important Deleted for CBSE Board 2024 Exams
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Question 9 Important Deleted for CBSE Board 2024 Exams
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Question 9 Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Misc 10 Important
Question 13 Important Deleted for CBSE Board 2024 Exams
Misc 14 Important
Misc 18 Important
Chapter 9 Class 11 Sequences and Series
Last updated at May 29, 2023 by Teachoo
Question 10 Find the sum to n terms of the series whose nth terms is given by (2n 1)2 Given an = (2n 1)2 =(2n)2 + (1)2 2(2n)(1) = 4n2 + 1 4n = 4n2 4n + 1 Sum of n terms is = 4((n(n+1)(2n+1))/6) 4 (n(n+1)/2) + n = n ("4" (n(n+1)(2n+1))/6 " 4" ( ( + 1))/2 " + 1" ) = n (2/3 " (n + 1)(2n + 1) 2(n + 1) + 1" ) = n ((2(n + 1)(2n + 1) 6(n + 1) + 3)/3) = /3 [(2n + 2)(2n + 1) 6n 6 + 3 ] = /3 [4n2 + 2n + 4n + 2 6n 3] = /3 [4n2 + 6n 6n 1] = /3 [4n2 1] = /3 [(2n)2 (1)2] = /3 [(2n 1) (2n + 1)] Hence, the required sum is /3 [(2n 1 ) (2n + 1)]