#14

Prove that if n is an odd positive integer, there exists a sequence of n consecutive integers whose sum is equal to n².
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#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 (2007-06-02 00:15:24)

#16

#18

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

#19

#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

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

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

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 (2009-04-15 09:48:04)

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 (2009-04-20 16:24:25)