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heres the problem
(a)Determine all nonnegative integers r such that it is possible for an infinite arithmetic sequence to contain exactly r terms that are integers. Prove your answer.
seems like an awefull lot of cases
(b)Determine all nonnegative integers r such that it is possible for an infinite geometric sequence to contain exactly r terms that are integers. Prove your answer.
same thing
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hi thedarktiger
Are there a lot of cases ? I'm not so sure.
(a) Let the first term be a and the difference be d.
Then the sequence is a, a+d, a + 2d, a + 3d, ......
If a and d are both integers, then every term will be an integer and we're not interested in those, as we want r to be a finite number.
So at least one of them must not be an integer.
If d is irrational then kd for all whole numbers, k, will be irrational. So you'll only get a + kd to be an integer for one value of a.
If d is a proper fraction (ie. not whole) with denominator b , then an integer will pop up every time k = a multiple of b. Once again, an infinite number of cases.
I think you can answer this question by considering cases like this.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Last edited by Olinguito (2015-01-17 07:56:41)
Bassaricyon neblina
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How do you take down a post
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Hi;
You mean delete this thread?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I just want to know how
Good. You can read.
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At the bottom of each post you should see a delete button.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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