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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#14**

Prove that if *n* is an odd positive integer, there exists a sequence of *n* consecutive integers whose sum is equal to *n*[sup]2[/sup].

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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

#13 I surrender....

Numbers are the essence of the Universe

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#15** This is an easy one.

Prove that any even power of any odd integer leaves a remainder of 1 when divided by 4.

*Last edited by JaneFairfax (2007-06-01 02:15:24)*

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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

This one is easier

*Last edited by Stanley_Marsh (2007-06-01 07:39:28)*

Numbers are the essence of the Universe

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

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**Stanley_Marsh****Member**- Registered: 2006-12-13
- Posts: 345

Yah , it comes from mutiplication law of mod

Numbers are the essence of the Universe

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#16**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#17** Here is another exercise I made up myself.

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#18**

*Last edited by JaneFairfax (2007-08-07 10:48:00)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

This result is important because I used it in proving Tonys question: http://www.mathsisfun.com/forum/viewtopic.php?id=7807

*Last edited by JaneFairfax (2007-08-11 03:10:04)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#19**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#20**

I just made this problem up myself; hopefully Ive got the maths correct.

*Last edited by JaneFairfax (2007-09-13 11:26:02)*

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**#21**

Consider the sequence

2, 5, 9, 12, 16, 19

The first term is 2 and successive terms are formed by alternately adding 3 and 4 to previous terms:

Prove that

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**Kurre****Member**- Registered: 2006-07-18
- Posts: 280

a more powerful result of 15:

which is two consecutive even integers, thus one can be written on the form 4k, and one on the form 4k+2, and thus k^{2n}-1 is divisible by 8, ie k^{2n} gives rest 1 when divided by 8.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,329

Hi Jane;

Answer for #21

Don't suppose anyone will ever see this but here goes:

The recurrence formula for your set of numbers is.

this has a characteristic equation of

x^3-x^2-x+1=0 which has roots of {-1,1,1}

This then is the general form of the solution:

we need to determine c1,c2, and c3 from the initial conditions

So the general solution is:

This agrees with your general solution at the bottom of problem 21

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

**bobbym**. looks like youve got the answer. Well done!

**My solution:**

Let

Then

and are APs with common difference 7 and first terms 2 and 5 respectively. ThusNote that

can be written as:The result follows.

*Last edited by JaneFairfax (2009-04-14 01:22:51)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,329

Hi Jane;

Thanks for showing me yours, I am not that happy with mine, what justification do I have for the first step? I haven't proven that a(n+3)=a(n+2)+a(n+1)-a(n) is the recursion for that sequence. Can you provide some rigor to my arguments?

bobbym

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Well, Im not that familiar with techniques on solving difference equations, but your solution looked right to me so I gave you full credit for it.

Im pretty sure your first step was correct.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,329

Hi Jane;

You say you are not very familiar with methods of solving difference equations but yet you came up with the idea of splitting the sequence into 2 coupled difference equations, each handling every other term. I salute you.

bobbym

*Last edited by bobbym (2009-04-14 11:48:04)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**JaneFairfax****Member**- Registered: 2007-02-23
- Posts: 6,868

Aw, thanks. Well, I thought up that problem myself, so for this particular problem I happen to have have my own solution handy.

At least, I discovered the following handy lemma myself:

If a sequence A has properties that alternate between its odd and even terms, let B be the subsequence of its odd-numbered terms and C the subsequence of its even-numbered terms. Find formulas for B and C separately. Then combine the formulas by rewriting A in the manner described above.

Id love to call this Janes lemma if no-one else has discovered it before.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 92,329

Hi Jane;

I discovered that about 10 years ago but didn't notice your method of combining them and promptly forgot about the whole idea. So I guess its yours and I will always remember it as Jane's lemma.

bobbym

*Last edited by bobbym (2009-04-19 18:24:25)*

**In mathematics, you don't understand things. You just get used to them.**

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