- #1
skate_nerd
- 174
- 0
I have a problem and honestly have no idea even where to start. I've been staring at it and thinking about it for over 24 hours...
Let \(u_n\) denote the \(n^{th}\) Fibonacci number. Without using the Binet formula for \(u_n\), prove the following for all natural numbers \(m\) and \(n\) with \(m\geq{2}\):
$$u_{m+n}=u_{m-1}u_n+u_mu_{n+1}$$
I have gone through a couple proofs regarding the Fibonacci numbers in this class before, but never one with two unknowns, and also never one where we weren't able to use the Binet formula. Without the Binet formula, what do I have to work off of? Do I need to write the derivation for the Binet formula within this whole proof?
Let \(u_n\) denote the \(n^{th}\) Fibonacci number. Without using the Binet formula for \(u_n\), prove the following for all natural numbers \(m\) and \(n\) with \(m\geq{2}\):
$$u_{m+n}=u_{m-1}u_n+u_mu_{n+1}$$
I have gone through a couple proofs regarding the Fibonacci numbers in this class before, but never one with two unknowns, and also never one where we weren't able to use the Binet formula. Without the Binet formula, what do I have to work off of? Do I need to write the derivation for the Binet formula within this whole proof?
Last edited:
