Compute $p_{n+1} = p^2 \frac{r_1^n-r_2^n}{r_1-r_2}, n \geq 0$

Question

I need to compute $$p_{n+1} = p^2 \frac{r_1^n-r_2^n}{r_1-r_2}, n \geq 0$$ where $p=\frac{1}{2}$, $r_1= \frac{1+\sqrt5}{4}= \frac{1}{2}\varphi$ and $r_2= \frac{1-\sqrt5}{4}$.

Using properties of Fibonnaci numbers I've found that $$p_{n+1}=2^{-n} F_n $$ with $F_n$ the $n$-th Fibonnaci number. Is there a way to find an explicite value for $p_{n+1}$ ? I've tried a geometric summation but I couldn't conclude too.

Thanks !

Answer 1

With the binomial theorem,

$$p_{n+1} = \frac{1}{4} \frac{ \sum_{k=0}^{n} \binom{n}{k} \frac{(\sqrt{5})^k}{4^n} - \sum_{k=0}^{n} \binom{n}{k} \frac{(-1)^k(\sqrt{5})^k}{4^n} }{ \frac{\sqrt{5}}{2}}$$

$$=\frac{ \sum_{k=0}^{n/2} \binom{n}{2k} \frac{(\sqrt{5})^{2k}}{4^n} + \sum_{k=0}^{n/2} \binom{n}{2k+1} \frac{(\sqrt{5})^{2k+1}}{4^n} - \sum_{k=0}^{n/2} \binom{n}{2k} \frac{(\sqrt{5})^{2k}}{4^n} + \sum_{k=0}^{n/2} \binom{n}{2k+1} \frac{(\sqrt{5})^{2k+1}}{4^n} }{ 2\sqrt{5}}$$ so $$p_{n+1}= \frac{ 2\sum_{k=0}^{n/2} \binom{n}{2k+1} \frac{(\sqrt{5})^{2k+1}}{4^n} }{ 2\sqrt{5}} = \frac{\sum_{k=0}^{n/2} \binom{n}{2k+1} (\sqrt{5})^{2k}}{4^n} = \frac{\sum_{k=0}^{n/2} \binom{n}{2k+1} (5)^{k}}{4^n}$$

EDIT: Note that $n$-even is assumed. If $n$ is odd, just replace $n/2$ with $(n-1)/2$.

Answer 2

The answer given by @DogeChan, while correct, is unnecessarily complex. The Fibonacci sequence is sufficiently well known that it provides a satisfactory solution. Your solution for $p_{n+1}$ is unfortunately incorrect. Specifically, for $p=1/2, r_1=\varphi/2, r2=(1-\varphi)/2$, I find that

$$ (r_1^n-r_2^n)/(r_1-r_2)=\frac{1}{2^{n-1}}\frac{\varphi^n-\psi^n}{\varphi-\psi}\\ $$

where $\psi=1=\varphi$.

Then we have

$$ p_{n+1}=\frac{F_n}{2^{n+1}}=\frac{1}{2^{n+1}}\lfloor\varphi^n/\sqrt{5}\ \rceil $$

where the brackets indicate the nearest integer (i..e., rounding), just in case you don't know the Fibonacci sequence offhand. I have checked all these solutions numerically.