Infinite Golden Sum

$\begin{aligned} S & = \varphi^n + \varphi^{n-1} + \varphi^{n-2} + \cdots & \small \color{#3D99F6} \text{where }\varphi \text{ is golden ratio.} \\ & = \varphi^n \color{#3D99F6} \sum_{k=0}^\infty \varphi^{-k} & \small \color{#3D99F6} \text{Sum of an infinite geometric progression} \\ & = \varphi^n \cdot \color{#3D99F6} \frac 1{1-\varphi^{-1}} \\ & = \varphi^n \cdot \frac {\color{#3D99F6}\varphi}{\varphi -1} & \small \color{#3D99F6} \text{Note that }\varphi^2 - \varphi - 1 = 0 \\ & = \varphi^n \cdot \frac {\color{#3D99F6}\varphi^2-1}{\varphi -1} & \small \color{#3D99F6} \implies \varphi = \varphi^2 - 1 \\ & = \varphi^n \cdot \frac {(\varphi-1)(\varphi + 1)}{\varphi -1} \\ & = \varphi^n \color{#3D99F6} (\varphi + 1) & \small \color{#3D99F6} \text{and }\varphi^2 = \varphi + 1 \\ & = \boxed{\varphi^{n+2}} \end{aligned}$

We have a infinite Geometric Progression with common ratio $\frac{1}{\phi}$ . This common ratio is smaller than $1$ , so we can use the infinite GP sum formula:

$S=\frac{a}{1-r}$ , where $S$ is the limit of the series, $a$ is the first term of the GP and $r$ is the common ratio. Substituting:

$S=\frac{\phi^n}{1-\frac{1}{\phi}}$

$S=\frac{\phi^n}{\frac{\phi-1}{\phi}}$

$S=\frac{\phi^{n+1}}{\phi-1}$

By the properties of the golden ratio we have that $\phi-1=\frac{1}{\phi}$ . Hence:

$S=\frac{\phi^{n+1}}{\frac{1}{\phi}}$

$S=\phi^{n+2}$

Then the final answer is $\boxed{\phi^{n+2}}$ .