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May 22nd, 2013, 12:22 PM   #1
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Kinda trivial, but kinda neat!

While gazing at and admiring the j-prime [1,2,2,3,5], I noticed that it was a very near miss to the fibonacci sequence 1,1,2,3,5 and that the sum of the first 4 members are the next fibonacci number 8 and the sum of all five is 13, the next one yet. This led me to conjecture that the sum of the first n fibonacci numbers is 1 less than the n+2nd fibonacci number.

So, the "sum" of the first fibonacci number is 1, which is indeed 1 less than the 3rd fibonacci number, ie 2.

So assume that the sum of the first n fibonacci numbers is if the n+2 fibonacci number minus 1.
That means that the sum of the first n+1 fibonacci numbers equals the the sum of the first n + the n+1st fibonacci number.

Ok, so the n+3rd fibonacci number is the sum of the n+1st + n+2nd fibonacci numbers. By assumption the n+2nd fibonacci number minus 1. So the sum of the the first n+1 fibonaccie numbers equals the n+1st fibonnaci number plus the n+2nd minus 1. Meanwhile, the n+3rd fibonacci number equals the n+1st plus n+2nd fibonacci number. So the sum of the first n+1 fibonacci numbers is one less than then n+3rd, ie the (n+1)+2nd fibonacci number, completing the induction.

I'm sure this is a long since known result, but it was fun running into it unexpectedly and working it out!
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May 22nd, 2013, 12:29 PM   #2
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Re: Kinda trivial, but kinda neat!

Indeed.

Usually now I just look up these sorts of sequences in the OEIS; it's faster than doing the quick induction.
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May 22nd, 2013, 12:33 PM   #3
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Re: Kinda trivial, but kinda neat!

Quote:
Originally Posted by CRGreathouse
Indeed.

Usually now I just look up these sorts of sequences in the OEIS; it's faster than doing the quick induction.
Ah yes, but some of us need a little practice!
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