Let P_n = _ k=0 ^ n-1 ( 2^k 129 ) . If P_n = 1 2^n , then the smallest positive integer n that satisfies this equation is:

Step-by-step solution

Given P_n = _ k=0 ^ n-1 ( 2^k 129 ) Using the standard identity for the product of cosines with angles in a geometric progression: _ k=0 ^ n-1 (2^k ) = (2^n ) 2^n Substitute = 129 : P_n = ( 2^n 129 ) 2^n ( 129 ) We are given that P_n = 1 2^n . Equating the two expressions: ( 2^n 129 ) 2^n ( 129 ) = 1 2^n ( 2^n 129 ) = ( 129 ) The general solution for x = y is x = m + (-1)^m y , where m is an integer. 2^n 129 = m + (-1)^m 129 Dividing by and multiplying by 129 : 2^n = 129m + (-1)^m Case 1: m is an even integer, say