The sum of the first 2n terms of the series 3(1)^2 - 2^2 + 3(3)^2 - 4^2 + 3(5)^2 - 6^2 + is 4300 . The value of n is

Step-by-step solution

Let the sum of the first 2n terms be S . Grouping the adjacent terms into n pairs, the k -th pair is given by: P_k = 3(2k-1)^2 - (2k)^2 P_k = 3(4k^2 - 4k + 1) - 4k^2 = 8k^2 - 12k + 3 The sum of the first n pairs is: S = _ k=1 ^n (8k^2 - 12k + 3) = 8 _ k=1 ^n k^2 - 12 _ k=1 ^n k + 3 _ k=1 ^n 1 S = 8 ( n(n+1)(2n+1) 6 ) - 12 ( n(n+1) 2 ) + 3n S = n 3 [ 4(2n^2 + 3n + 1) - 18(n+1) + 9 ] S = n 3 [ 8n^2 + 12n + 4 - 18n - 18 + 9 ] = n 3 [ 8n^2 - 6n - 5 ] We are given S = 4300 , so: n 3 [ 8n^2 - 6n - 5 ] = 4300 n(8n^2 - 6n