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### Fibonacci Trick

```
Date: 06/01/99 at 18:16:10
From: Stephanie Roberts
Subject: Fibonacci Sequence

I am interested in why the Fibonacci trick works. How do f(n-1),
f(n+1), and f(n)^2 relate to each other? I am not a good math student
at all but the Fibonacci sequence has really interested me. I have
been working for a while on a way to solve this but I haven't gotten
see if it helps to understand the ways the Fibonacci sequence shows up
in nature, etc.

Thank you so much.
Stephanie
```

```
Date: 06/01/99 at 18:46:55
From: Doctor Anthony
Subject: Re: Fibonacci Trick

For the Fibonacci sequence show:

[F(n+1) * F(n-1)] - [F(n)]^2 = (-1)^n

The expression for the nth term of the Fibonacci sequence is:

1      [1+sqrt(5)]^n        1      [1-sqrt(5)]^n
F(n) = ------- * -------------  -  ------- * -------------
sqrt(5)        2^n          sqrt(5)        2^n

I will write this in the form

1
F(n) = ------- * [P^n - Q^n]
sqrt(5)

1+sqrt(5)                 1-sqrt(5)
where  P = ---------     and     Q = ---------
2                         2

Note that  P+Q = 1
P-Q = sqrt(5)
PQ = 1/4 - 5/4 = -1

We are asked to show that  [F(n+1) * F(n-1)] - [F(n)]^2 = (-1)^n

[F(n)]^2 = (1/5)*[P^n - Q^n]^2

= (1/5)*[P^(2n) + Q^(2n) - 2*P^n*Q^n]    ..............(1)

F(n+1) * F(n-1) = (1/5)*[P^(n+1) - Q^(n+1)]*[P^(n-1) - Q^(n-1)]

= (1/5)*[P^(2n) + Q^(2n) - P^(n+1)*Q^(n-1)
- P(n-1)*Q^(n+1)]              ..............(2)

subtracting (1) from (2) we get:

(1/5)*[-P^(n+1)*Q^(n-1) - P^(n-1)*Q^(n+1) + 2*P^n*Q^n]

= (1/5)*[-P^(n-1)*Q^(n-1)*(P^2 + Q^2) + 2*P^n*Q^n]

= (1/5)*P^(n-1)*Q^(n-1)*[-(P^2 + Q^2) + 2*P*Q]

= -(1/5)*P^(n-1)*Q^(n-1)*[P^2 - 2*P*Q + Q^2]

= -(1/5)*P^(n-1)*Q^(n-1)*(P-Q)^2

but (P-Q)^2 = 5, so we obtain

= -P^(n-1)*Q^(n-1)

= -(PQ)^(n-1)

but PQ = -1 and so the difference between (1) and (2) reduces to:

= -(-1)^(n-1)

= (-1)^n

and this completes the proof.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Fibonacci Sequence/Golden Ratio

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