This problem arose when looking into the area of a dignomonic tiling. I found an identity for an arbitrary number, call it $\Phi$, that is totally independent of the tiling itself. The result is given as follows:
$$ 1+(\Phi^2-1)\sum_{k=0}^{m-1}\Phi^{2k}= \Phi^{2m} $$
Here, $m$ is the number of gnomonic pairs. To put this perspective, the first term on the LHS represents the area of the tiling seed, $(\Phi^2-1)$ represents the area of the initial gnomonic pair, while the summation represents the area growth of successive gnomon pairs. It’s interesting that while this identity was derived for positive $\Phi$, it is true for negative and complex $\Phi$ as well. Dignomonic tiling is dependent upon two parameters, $\phi_r$ and $\phi_s$ with the seed tile being $1/\phi_s\times 1$ and the growth rate is $\Phi=\phi_r \phi_s$. To obtain the area, just multiply the above equation by $1/\phi_s$, i.e., the area of the seed. You can find out more about dignomonic tiling here or here: M.J. Gazalé, Gnomon: From Pharaohs to Fractals. Princeton University Press (1999).
The above identity is anything but obvious and I was wondering how to resolve it. After working up this question, I had an epiphany. Actually, two. See my answers below.
The enlightened way: Beginning with
$$ 1+(\Phi^2-1)\sum_{k=0}^{m-1}\Phi^{2k}=\Phi^{2m} $$
It becomes apparent that the LHS can be expressed as a recurrence formula, that is,
$$ f_k=f_{k-1}+(\Phi^2-1) \Phi^{2k-2}, \quad f_0=1 $$
By induction then,
$$ \begin{align} &f_1=f_0+(\Phi^2-1) \Phi^0=1+(\Phi^2-1)= \Phi^2\\ &f_2= f_1+(\Phi^2-1) \Phi^2=\Phi^2+(\Phi^2-1) \Phi^2=\Phi^4\\ &\vdots\\ &f_k=\Phi^{2k} \end{align} $$ The easier way: Here we utilize the partial sum equation
$$ \sum_{k=1}^n x^k=\frac{x}{x-1}(x^n-1) $$
Then
$$ \begin{align} &1+(\Phi^2-1)\sum_{k=0}^{m-1}\Phi^{2k}= \Phi^{2m}\\ &1+(\Phi^2-1)+ (\Phi^2-1)\sum_{k=1}^{m-1}\Phi^{2k}= \Phi^{2m}\\ &\Phi^2+ (\Phi^2-1)\frac{\Phi^2}{\Phi^2-1}( \Phi^{2(m-1)}-1) =\Phi^{2m}\\ &\Phi^{2m}=\Phi^{2m}\\ \end{align} $$