Difference between revisions of "2018 IMO Problems/Problem 2"
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− | We find at least one series of real numbers for <math>n = 3,</math> for each <math>n = 3k</math> and we prove that if <math>n = 3k | + | We find at least one series of real numbers for <math>n = 3,</math> for each <math>n = 3k</math> and we prove that if <math>n = 3k \pm 1,</math> then the series does not exist. |
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+ | <i><b>Case 1</b></i> Let <math>n = 3.</math> We get system of equations | ||
+ | <cmath> | ||
+ | |||
+ | We subtract the first equation from the second and get: <cmath>a_2 (a_3 – a_1) = (a_1 – a_3).</cmath> | ||
+ | So <math>a_2 = – 1 \implies a_1 = 2, a_3 = – 1.</math> | ||
+ | |||
+ | <i><b>Case 1'</b></i> Let <math>n = 3k, k={1,2,...}.</math> The sequence <math>{ 2, – 1, – 1, ..., 2, – 1, – 1}</math> is the desired sequence. |
Revision as of 02:03, 16 August 2022
Find all numbers for which there exists real numbers
satisfying
and
for
Solution
We find at least one series of real numbers for for each
and we prove that if
then the series does not exist.
Case 1 Let We get system of equations
We subtract the first equation from the second and get:
So
Case 1' Let The sequence
is the desired sequence.