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Ex 9.4
Ex 9.4, 2 Important Deleted for CBSE Board 2023 Exams
Ex 9.4, 3 Deleted for CBSE Board 2023 Exams
Ex 9.4.4 Important Deleted for CBSE Board 2023 Exams
Ex 9.4, 5 Deleted for CBSE Board 2023 Exams
Ex 9.4, 6 Important Deleted for CBSE Board 2023 Exams
Ex 9.4, 7 Important Deleted for CBSE Board 2023 Exams
Ex 9.4, 8 Deleted for CBSE Board 2023 Exams
Ex 9.4, 9 Important Deleted for CBSE Board 2023 Exams
Ex 9.4, 10 Deleted for CBSE Board 2023 Exams You are here
Ex 9.4
Last updated at March 16, 2023 by Teachoo
Ex 9.4, 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)]