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 |

Here's where it gets interesting. If you square both numbers, you will see that 5

^{2}= 25, and 8

^{2}= 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. 13

^{2}= 169, and 21

^{2}= 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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