Mock AIME 1 2006-2007 Problems/Problem 9
Let be a geometric sequence of complex numbers with and , and let denote the infinite sum . If the sum of all possible distinct values of is where and are relatively prime positive integers, compute the sum of the positive prime factors of .
Let be a geometric sequence for with and . Let denote the infinite sum: . If the sum of all distinct values of is where and are relatively prime positive integers, then compute the sum of the positive prime factors of .
Then the sum has value . Different choices of clearly lead to different values for , so we don't need to worry about the distinctness condition in the problem. Then the value we want is . Now, recall that if are the th roots of unity then for any integer , is 0 unless in which case it is 1. Thus this simplifies to
We seek , or the negative of the coefficient of divided by the coefficient of , which is and .
Therefore the answer is .
To answer the original problem statement, let the common geometric be and denotes a term in the sequence. The term can be represented as and this expression has to be set equal to one. By simplifying, we get where the polynomial can have ten possible values or roots. Now, we take a look at the needed infinite geometric sum. The expression can be expressed as or even more better, . However, the question is asking for the sum of the distinct roots of the polynomial, and as stated before, is a tenth-degree polynomial and it had ten different roots. Therefore, our desired sum is . We can keenly set to get rid of the fractions and express it in a different variable. This leads us to or . Plugging it back to our original equation , we get . By Vieta's, the sum of the roots is which is equal to . However, notice that is only needed, so we have to find the coefficient of the tenth-exponent. By the Binomial Theorem, this is equal to and by simplifying and factoring , we get . Thus, can be rewritten as , and the final answer is .