Golden base

Consider that $\phi$ is one of the roots of $x^2-x-1=0$

So, $\phi^2=\phi+1$

$\phi^3=\phi^2+\phi=2\phi+1$

$\phi^3-1=2\phi$

$1-\phi^{-3}=2\phi^{-2}$

$\dfrac{1}{1-\phi^{-3}} = \dfrac{\phi^2}{2}$

$1+\phi^{-3}+\phi^{-6}+\phi^{-9}+\cdots=\dfrac{\phi^2}{2}$

$\phi^{-1}+\phi^{-4}+\phi^{-7}+\cdots=\dfrac{\phi}{2}$

hence the answer.

Let $\phi =\dfrac { 1+\sqrt { 5 } }{ 2 }$

Then we consider the following family of infinite sums

$\displaystyle \sum _{ k=0 }^{ \infty }{ { \phi }^{ -\left( nk+1 \right) } } =\dfrac { { \phi }^{ n } }{ { \phi }^{ n+1 }-1 }$

so that we seek integer $n$ such that

$\dfrac { { \phi }^{ n } }{ { \phi }^{ n+1 }-1 } =\dfrac { \phi }{ 2 }$

or

${ \phi }^{ n-2 }({ \phi }^{ 2 }-1)-1=0$

Since $\phi$ satisfies

${ \phi }^{ 2 }-\phi -1=\phi (\phi -1)-1=0$

and

${ \phi }^{ 2 }-2=\phi -1$

we end up with $n=3$