Let S_n = _ k=1 ^ n ( (2k-1) 12 ) ( (2k+1) 12 ) . If S_n = 2 3 - 6 , then the sum of the first three positive integer values of n for which this holds is

Step-by-step solution

The general term of the series is T_k = ( (2k-1) 12 ) ( (2k+1) 12 ) = 1 ( (2k-1) 12 ) ( (2k+1) 12 ) . Notice that the difference between the angles is (2k+1) 12 - (2k-1) 12 = 2 12 = 6 . Multiplying and dividing by ( 6 ) = 1 2 , we get: T_k = 2 ( (2k+1) 12 - (2k-1) 12 ) ( (2k-1) 12 ) ( (2k+1) 12 ) = 2 [ ( (2k+1) 12 ) - ( (2k-1) 12 ) ] Summing from k=1 to n , the series telescopes: S_n = _ k=1 ^ n T_k = 2 [ ( 3 12 ) - ( 12 ) + ( 5 12 ) - ( 3 12 ) + + ( (2n+1) 12 ) - ( (2n-1) 12 ) ] S_n = 2 [ ( (2n+1) 12 ) - ( 12 ) ]