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# Geometry/Induction

Question
Hello,

Use induction to show the following Fibonacci numbers:
F(n-1) * F(n+2) = (F(n) * Fn+1) + (-1)^n for every integer n >= 2

Thanks

This does give a standard Fibonacci sequence if we started with
F(0)=1, F(1)=1, and F(2)=2.  Starting with F(4), we get 3, 5, 8, 13, 21, 34, 55, 89, 144 ...

However, there is nothing here that says we need to start this way.

If we started with F(0)=1, F(1)=1, and F(2)=1, we would get a series where
F(n) = 1 when n is odd and F(n)=n/2 when n is even.

If we started with F(0)=1, F(1)=-1, and F(2)=1, starting at the 7th term,
all terms thereafter are undefined.

If we started with F(0)=1, F(1)=1, and F(2)=3, we would get a totally different series.
It would be 4, 11, 15, 41, 56, 153, 209, 571, 780, 2131, 2911, 7953, 10864 ...
I have no clue what that series looks like right now, but could always look into it.
It looks to me like a twisted form of the Fibonacci sequence.

If we started with F(0)=1, F(1)=-1, and F(2)=2, we would get the same result as the first one on the even terms and -1 for the odd terms.

If we started with F(0)=1, F(1)=-2, and F(2)=3, we would get the Fibonacci series with all of the even terms negative.

If we started with F(0)=1, F(1)=2, and F(2)=2, we would get a series with increasing 2's in the denominator.  That is, the series would be 1, 2, 2, 5, 4 1/2, 11 3/4, 10 3/8, 27 5/16,
24 1/32, 63 23/64, 55 91/128, 146 237/256, 129 89/512, 340 703/1024, 299 5/2048, etc.
I don't seen any easily detected pattern here, but it could be found with linear regression
and is suspected to also be a resemblance of the Fibonacci series.

Geometry

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