TimeFan

The sum of the first eight Fibonacci numbers is equal to the tenth Fibonacci number minus one.

Exercise: Prove the generalized form of this claim.

Generalised (because I'm British) claim:
The sum of any 8 consecutive Fibonacci numbers n1 - n8 is equal to the Fibonacci number n10 - n2.

I could notate this more clearly if only I could be bothered to render it in LaTeX :P

Proof:
Lets call the first in our set of consecutive numbers a, and the second b. Then we have the sequence:

The sum of the first 8 of these is 21a+33b.

The difference then between sum(n1:n8) and n10 is b, which is exactly n2, and thus sum(n1:n8) is n10 - n2

The other generalization:http://dl.dropbox.com/u/4386762/fibonacci.pdf

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It Will Control You

Lovely stuff, I wish I had bothered to lay mine out like that... :)

I wonder which of the two generalisations Ben had in mind?

William's is the one I had in mind, but both answers are acceptable.

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Not so fast, Unwinder! If you hold the toad, you have to join!

Of course, they both generalize to the claim that the sum of the mth to nth Fibonnaci numbers is the (n+2)th Fibonacci number minus the (m+1)th.