Difference between revisions of "1967 IMO Problems/Problem 5"
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| + | ==Problem== | ||
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Let <math>a_1,\ldots,a_8</math> be reals, not all equal to zero. Let <cmath>c_n = \sum^8_{k=1} a^n_k</cmath> for <math>n=1,2,3,\ldots</math>. Given that among the numbers of the sequence <math>(c_n)</math>, there are infinitely many equal to zero, determine all the values of <math>n</math> for which <math>c_n = 0.</math> | Let <math>a_1,\ldots,a_8</math> be reals, not all equal to zero. Let <cmath>c_n = \sum^8_{k=1} a^n_k</cmath> for <math>n=1,2,3,\ldots</math>. Given that among the numbers of the sequence <math>(c_n)</math>, there are infinitely many equal to zero, determine all the values of <math>n</math> for which <math>c_n = 0.</math> | ||
==Solution== | ==Solution== | ||
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| + | <math>c_n</math> must be zero for all odd <math>n</math>. | ||
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| + | Proof: | ||
| + | WLOG suppose that <math>a_1 \geq a_2 \geq ... \geq a_8</math>. If <math>a_1+a_8 > 0</math> then for sufficiently high odd <math>n</math>, <math>c_n</math> will be dominated by <math>a_1</math> alone i.e. it will always be positive. Similarly if <math>a_1+a_8 < 0</math>; hence <math>a_1=-a_8</math>. Now for odd <math>n</math> these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd <math>n</math>. Since some <math>a_i</math> is nonzero <math>c_n > 0</math> for even <math>n</math>. | ||
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| + | The above solution was written by Fiachra and can be found here: [https://aops.com/community/p138611] | ||
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Latest revision as of 19:23, 10 November 2024
Problem
Let 
be reals, not all equal to zero. Let ![\[c_n = \sum^8_{k=1} a^n_k\]](http://latex.artofproblemsolving.com/b/c/6/bc62323075b4e2015085cc7aa2e3364679d89ced.png)
for 
. Given that among the numbers of the sequence 
, there are infinitely many equal to zero, determine all the values of 
for which 
Solution
must be zero for all odd .
Proof:
WLOG suppose that 
. If 
then for sufficiently high odd , will be dominated by 
alone i.e. it will always be positive. Similarly if 
; hence 
. Now for odd these terms cancel, so we can repeat for the remaining values. Now all the terms cancel for all odd . Since some 
is nonzero 
for even .
The above solution was written by Fiachra and can be found here: [1]
Problem 5 on this (https://artofproblemsolving.com/wiki/index.php/1967_IMO_Problems) page is equivalent to this since the only difference is that they are phrased differently.
See Also
| 1967 IMO (Problems) • Resources | ||
| Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
| All IMO Problems and Solutions | ||
