#238 maths!
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Author:  Spider [ December 24th, 2010 09:23:54 am ]
Post subject:  #238 maths!

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

Lets call the first in our set of consecutive numbers a, and the second b. Then we have the sequence:
n1  n2  n3    n4     n5      n6      n7      n8       n9        n10
a   b   a+b   a+2b   2a+3b   3a+5b   5a+8b   8a+13b   13a+21b   21a+34b

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

Author:  william [ December 26th, 2010 07:55:18 am ]
Post subject:  Re: #238 maths!

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

Author:  Spider [ January 4th, 2011 14:53:52 pm ]
Post subject:  Re: #238 maths!

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

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

Author:  Diane Heaton [ January 4th, 2011 22:33:43 pm ]
Post subject:  Re: #238 maths!

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

Author:  Token [ January 20th, 2011 15:50:47 pm ]
Post subject:  Re: #238 maths!

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.

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