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How can I prove by induction that this is a closed form of the Fibonacci sequence?

$$F_n=\frac1{\sqrt5}\left(\frac{1+\sqrt5}2\right)^{n+1}-\frac1{\sqrt5}\left(\frac{1-\sqrt5}2\right)^{n+1}$$

I've never seen before the $n+1$ exponent

Brian M. Scott
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kiwi
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    Are you asking why the formula for $F_{\color{red}n}$ has $\color{red}{n+1}$ in the exponents? Some people use a non-standard indexing of the Fibonacci sequence, so that for them $F_0=1$, not $0$. That means that what they call $F_n$ is what I call $F_{n+1}$ and does need $n+1$ in the exponents. – Brian M. Scott Apr 06 '16 at 19:37

2 Answers2

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How to write a proof by induction.

Step 1: Show that it is true in the base case $F(0) = \frac{1}{\sqrt5}\phi - \frac{1}{\sqrt5}\phi'$ where $\phi = \dfrac{1+\sqrt5}{2}$ and $\phi' = \dfrac{1-\sqrt5}{2}$

Step 2: State the inductive hypothesis: "Assume that $F(k) = \frac{1}{\sqrt5}\phi^{k+1} - \frac{1}{\sqrt5}\phi'^{k+1}$ for all $k < n$"

Step 3: show that $F(n) = F(n-1) + F(n-2) = \frac{1}{\sqrt5}\phi^{n+1} - \frac{1}{\sqrt5}\phi'^{n+1}$

You get to use your hypothesis, i.e. $F(n-1) = \frac{1}{\sqrt5}\phi^n - \frac{1}{\sqrt5}\phi'^n$ and $F(n-2) =$ ... to help you.

Regarding the exponents, treat them like you would treat any other exponent. $(x)(x^n) = x^{n+1}$ They follow their algebraic rules, and just stick to the rules.

Doug M
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First of all, I think you have a mistake, I'm pretty sure the exponent is $n$ and not $n+1$.

Second of all, here's a hint: Let's write: $\frac{1}{\sqrt{5}} (\frac{1+\sqrt{5}}{2})=\phi_1$ and $\frac{1}{\sqrt{5}} (\frac{1-\sqrt{5}}{2})=\phi_2$

Note that $\phi_1$ and $\phi_2$ are the solutions to the following equation: $x^2-x-1$ And therefore $\phi_1^2-\phi_1-1=0$

So we get that $\phi_1+1=\phi_1^2$ And in the same way $\phi_2+1=\phi_2^2$

Now try to do the induction by using this somewhere along the way.

nono
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