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Thursday, January 9, 2014

Strange and Wonderful

Behold, the beginning of the Fibonacci sequence:

n        1     2     3     4     5     6     7     8     9     10     11     12     13     14     15 ...

F(n)   1     1     2     3     5     8    13    21    34   55    89    144    233   377   610 ...


Photo credit:
www.theoremoftheday.org
Now, here's something you may not know. Pick any two consecutive Fibonacci numbers.... let's take numbers 5 and 6. You will find the fifth Fibonacci number listed as F(5), and it's actually the number five! F(6) is 8.

Here's where it gets interesting. If you square both numbers, you will see that 52 = 25, and 82 = 64. Now add those up and you get 25 + 64 = 89. This is another Fibonacci number! But wait- there's more. WHICH Fibonacci number is it? It's the 11th....as in... 5+6 = 11. Hmm...let's try that one again.

Let's pick two consecutive Fibonacci numbers. We'll pick Fibonacci number 7 (13) and Fibonacci number 8 (21). Now, square those numbers. 132 = 169, and 212 = 441. Add those together (169 + 441) and you get ... 610 - another Fibonacci number! And which one is it? Remember, we chose F(7) and F(8).... 7 + 8 = 15, and the 15th Fibonacci number is.... 610!

What does this look like mathematically? Well, perhaps it's more than you came here for, but here goes:

F(n)2 + F(n+1)2 = F(n + [n+1])

When you square and then sum consecutive Fibonacci numbers, the answer you get is the Fibonacci number at the location that is the sum of the two original locations.

Whew!

Wanna see if you 'get' it? The 29th Fibonacci number is 514,229.

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