Maths Olympiad Prep

Library / HMMT February /764 of 1302

HMMT February

Algebra Difficulty 5.3 AIME, harder Find the answer

Let z0+z1+z2+⋯z_{0}+z_{1}+z_{2}+\cdots be an infinite complex geometric series such that z0=1z_{0}=1 and z2013=120132013z_{2013}=\frac{1}{2013^{2013}}. Find the sum of all possible sums of this series.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

Clearly, the possible common ratios are the 2013 roots r1,r2,…,r2013r_{1}, r_{2}, \ldots, r_{2013} of the equation r2013=120132013r^{2013}=\frac{1}{2013^{2013}}. We want the sum of the values of xn=11−rnx_{n}=\frac{1}{1-r_{n}}, so we consider the polynomial whose roots are x1,x2,…,x2013x_{1}, x_{2}, \ldots, x_{2013}. It is easy to see that (1−1xn)2013=120132013\left(1-\frac{1}{x_{n}}\right)^{2013}=\frac{1}{2013^{2013}}, so it follows that the xnx_{n} are the roots of the polynomial equation 120132013x2013−(x−1)2013=0\frac{1}{2013^{2013}} x^{2013}-(x-1)^{2013}=0. The leading coefficient of this polynomial is 120132013−1\frac{1}{2013^{2013}}-1, and it follows easily from the Binomial Theorem that the next coefficient is 2013, so our answer is, by Vieta's Formulae, −2013120132013−1=2013201420132013−1-\frac{2013}{\frac{1}{2013^{2013}}-1}=\frac{2013^{2014}}{2013^{2013}-1}

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: Omni-MATH, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.