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

]]>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.

]]>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

]]>Im pretty sure your first step was correct.

]]>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

]]>**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.

]]>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

]]>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.]]>

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

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

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

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