Let z = ( 2 3 ) + i ( 2 3 ) . If _ k =1 ^ n (z^ k + z ^ k )^3 = 53 , then the value of n is

Step-by-step solution

We know that z = ( 2 3 ) + i ( 2 3 ) = , where is the complex cube root of unity. Then z = ^2 . The general term of the sum is T_k = ( ^k + ^ 2k )^3 . If k is not a multiple of 3, ^k + ^ 2k = + ^2 = -1 , so T_k = (-1)^3 = -1 . If k is a multiple of 3, ^k + ^ 2k = 1 + 1 = 2 , so T_k = (2)^3 = 8 . The sequence of terms T₁, T₂, T₃, is -1, -1, 8, -1, -1, 8, The sum of any three consecutive terms (one full period) is -1 - 1 + 8 = 6 . Let n = 3m + r , where r 0, 1, 2 . Sum of n terms = 6m + _ k=1 ^r T_k = 53 . If we take