HMMT 二月 2013 · 冲刺赛 · 第 28 题
HMMT February 2013 — Guts Round — Problem 28
题目详情
- [ 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#
英文原题
[ 17 ] Let z 0 + z 1 + z 2 + · · · be an infinite complex geometric series such that z 0 = 1 and z 2013 = 1
2013 2013 .
Find the sum of all possible sums of this series.
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HMMT 2013, 16 FEBRUARY 2013 — GUTS ROUND
Organization Team Team ID#
解析
英文解析
- [ 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 − rn
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 nx 2013
1 2013 2013 nare the roots of the polynomial equation x − ( x − 1) = 0. The leading coefficient of this
2013
2013
polynomial is − 1, and it follows easily from the Binomial Theorem that the next coefficient is 1
2013
2013
2013, so our answer is, by Vieta’s Formulae,
2014
2013 2013 − = .
20131
2013 − 1 − 1
2013
2013