In article
<garscosi-35F51E.15150423122007@[EMAIL PROTECTED]
>, Gary
S. Simon <garscosi@[EMAIL PROTECTED]
> wrote:
> Riding the subway home on the last leg of our return from the Staten
> Island Zoo yesterday, my seven year old son Dan asked me to do the
> following (Dan doesn't know much terminology, so I'm including it
> parenthetically):
>
> 1. Pick a Fibonacci number (Fn)
> 2. Square it
> 3. Square the Fibonacci "two before the one you picked" (F(n-2))
> 4. Subtract it from the first square
> 5. You get the Fibonacci when you add the other two (F(2n-2)).
[...]
Changing indices for convenience, you want
F(2n) = F(n + 1)^2 - F(n - 1)^2.
Brian has shown you one way; here is another:
Let A be the matrix
1 1
1 0.
Let U(n) be the matrix
F(n + 1) F(n)
F(n) F(n - 1)
It is not too hard to see that A^n = U(n).
Now, A^(2n) = A^n * A^n. So, U(n)*U(n) = U(2n). Looking at the (2,2)
entries gives F(n)^2 + F(n - 1)^2 = F(2n - 1).
Thus F(2n + 1) = F(n + 1)^2 + F(n)^2. On the other hand,
F(2n + 1) = F(2n - 1) + F(2n) = F(n)^2 + F(n - 1)^2 + F(2n).
Equating the two gives the desired result.
--
Paul Sperry
Columbia, SC (USA)
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