Re: twin prime conjecture

OK, here's the solution to one more of your problems (2 more to go):

1. Prove that F_1 + F_2 + F_3 + ... + F_n = F_(n + 2) - 1

(a) F(4) - 1 = F(3) + F(2) - 1 = F(2) + F(1) + F(2) - 1 = F(2) + F(2)
= F(2) + F(1) + F(0) = F(2) + F(1)

(b) Assume that this is true up to n so that:

F(n) - 1 = F(n-2) + ... + F(1)

(c) Then F(n+1) - 1 = F(n) + F(n-1) - 1

(d) Applying (b) gives us:

F(n+1) - 1 = F(n-1) + F(n-2) + ... F(1)

(e) Using mathematical induction, we are done.

QED

-Larry


Gerry Myerson wrote:
In article <1160544899.925138.289320@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
"larry.freeman@xxxxxxxxx" <larry.freeman@xxxxxxxxx> wrote:

Good work on 1, 2, 3, 4.

5. The Fibonacci numbers are given by F_0 = 0, F_1 = 1, and
F_n = F_(n - 1) + F_(n - 2) for n = 2, 3, ... (so they go
0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, ...). Let phi be the
number (1 + sqrt 5) / 2. Prove that
phi^(n - 1) < F_(n + 1) < phi^n if n > 1.

In other words, prove Binet's Formula.

No. Well, I suppose you could first prove Binet, and then use it
to solve the problem, but I reckon you should be able to do the
problem without reference to Binet.

--
Gerry Myerson (gerry@xxxxxxxxxxxxxxx) (i -> u for email)

.

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