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My initial prompt is as follows:

For $F_{0}=1$, $F_{1}=1$, and for $n\geq 1$, $F_{n+1}=F_{n}+F_{n-1}$. Prove for all $n\in \mathbb{N}$:

$$F_{n-1}=\frac{1}{\sqrt{5}}\left(\left(\frac{1+\sqrt5}{2}\right)^n-\left(\frac{1-\sqrt5}{2}\right)^n\right)$$

Which, to my understanding, is Binet's Formula.

I came up with a proof strategy similar to that described here. But as you can see, it is considering the case of $F_n$, not $F_{n-1}$.

So, my question is: does the same strategy hold for $F_{n-1}$? If not, how is it different?

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marked as duplicate by J. M., Amzoti, Alexander Gruber Jun 3 '13 at 3:10

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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@Whoever is voting to close: Why? This is a perfectly understandable question. –  anorton May 28 '13 at 22:54

2 Answers 2

up vote 5 down vote accepted

The problem that you’ve been given uses a less-standard indexing of the Fibonacci numbers. The usual definition has initial values $F_0=0$ and $F_1=1$; your $F_n$ is the usual $F_{n+1}$, and the usual $F_n$ is your $F_{n-1}$. Thus, where your link has $$\varphi^a=F(a)\varphi+F(a-1)\;,$$ you’ll need to write $$\varphi^a=F_{a-1}\varphi+F_{a-2}\;.$$ As a quick check, when $a=2$ that gives you $\varphi^2=F_1\varphi+F_0=\varphi+1$, which you can see from the link is correct.

(I’m assuming here that your proof really does follow pretty much the pattern in the link. You can also just do a straight induction: the induction step is the same whichever indexing you use.)

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It's impossible to tell without reading your proof, but induction (as a principle) will work regardless of whether you consider $n$ or $n-1$. The only difference is the indexing of the base case, but with the shift $n \mapsto n-1$ you needn't worry about introducing a "gap" at all.

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