HMMT 二月 2013 · 冲刺赛 · 第 28 题

HMMT February 2013 — Guts Round — Problem 28

[ 17 ] Let z + z + z + · · · be an infinite complex geometric series such that z = 1 and z = . 0 1 2 0 2013 2013 2013 Find the sum of all possible sums of this series. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND Organization Team Team ID#

解析

[ 17 ] Let z + z + z + · · · be an infinite complex geometric series such that z = 1 and z = . 0 1 2 0 2013 2013 2013 Find the sum of all possible sums of this series. 2014 2013 Answer: Clearly, the possible common ratios are the 2013 roots r , r , . . . , r of the 2013 1 2 2013 2013 − 1 1 1 2013 equation r = . We want the sum of the values of x = , so we consider the polynomial 2013 n 2013 1 − r n 1 1 2013 whose roots are x , x , . . . , x . It is easy to see that (1 − ) = , so it follows that the x 1 2 2013 2013 n x 2013 n 1 2013 2013 are the roots of the polynomial equation x − ( x − 1) = 0. The leading coefficient of this 2013 2013 1 polynomial is − 1, and it follows easily from the Binomial Theorem that the next coefficient is 2013 2013 2013, so our answer is, by Vieta’s Formulae, 2014 2013 2013 − = . 1 2013 2013 − 1 − 1 2013 2013