Question 10 - Finding sum from nth number - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Finding sum from nth number
Last updated at April 16, 2024 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)]