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 Post subject: #238 maths!Posted: December 24th, 2010 09:23:54 am
 Yorkster

Joined: January 29th, 2007 14:50:51 pm
Posts: 117
Location: Leicester, England
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:
Code:
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

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 Post subject: Re: #238 maths!Posted: December 26th, 2010 07:55:18 am
 Aoristocrat

Joined: November 10th, 2006 00:24:13 am
Posts: 853
Location: I'll never tell! (North Carolina)
The other generalization:http://dl.dropbox.com/u/4386762/fibonacci.pdf

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 Post subject: Re: #238 maths!Posted: January 4th, 2011 14:53:52 pm
 Yorkster

Joined: January 29th, 2007 14:50:51 pm
Posts: 117
Location: Leicester, England
Lovely stuff, I wish I had bothered to lay mine out like that... :)

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

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 Post subject: Re: #238 maths!Posted: January 4th, 2011 22:33:43 pm
 Factitious

Joined: April 3rd, 2006 00:57:16 am
Posts: 1294
William's is the one I had in mind, but both answers are acceptable.

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 Post subject: Re: #238 maths!Posted: January 20th, 2011 15:50:47 pm
 Fortnight

Joined: January 30th, 2008 11:28:07 am
Posts: 17
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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