Show that $\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10}$ 
Show that$$\sum_{k=1}^{\infty}\frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{15-\sqrt{5}}{10},$$
  where $F_n$ is a Fibonacci number and $L_n$ is a Lucas number.$^1$

Motivation: For example, when calculating Millin series
$$
\sum_{n=1}^{\infty} \frac{1}{F_{2^n}}=\frac{7-\sqrt{5}}{2},
$$
we can show that
$$
\sum_{k=1}^n \frac{1}{F_{2^k}} =3-\frac{F_{2^n-1}}{F_{2^n}}.
$$
and finish the calculation by sending $n$ to $\infty$.
Also, when proving
$$
\sum_{n=1}^{\infty}\frac{F_{2^{n-1}}}{L_{2^n}+1}=\frac{1}{\sqrt{5}},
$$
solution first shows that
$$
\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}+1}=\frac{F_{2^n}}{L_{2^n}+1}
$$
After deducting such formulas, it is easy to calculate limits. Then, how to find a formula about $$\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}-2}=?$$
My attempt: Since
$$
\frac{15-\sqrt{5}}{10} = \frac{3}{2}-\frac{1}{2}\cdot\frac{1}{\sqrt{5}},
$$
I made a conjecture
$$
\sum_{k=1}^n \frac{F_{2^{k-1}}}{L_{2^k}-2}=\frac{3}{2}-\frac{1}{2}\frac{F_{2^n}}{L_{2^n}-2}.
$$
For $n=1$,
$$\frac{F_1}{L_2-2}=1=\frac{3}{2}-\frac{1}{2}=\frac{3}{2}-\frac{1}{2}\frac{F_2}{L_2-2}.$$
Suppose that the conjecture holds for some $n$. Then
\begin{align}
\sum_{k=1}^{n+1} \frac{F_{2^{k-1}}}{L_{2^k}-2}&=\frac{3}{2}-\frac{1}{2}\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}\\
&=\dots?
\end{align}

My question: How to complete the induction step? If the conjecture is wrong, then how to find other ways?


$^1$ Elementary Problems and Solutions (May 2015). Fibonacci Quarterly. Volume 53. Number 2.

Using 
$$F_{2N}=F_NL_N$$
$$L_{2N}=L_N^2-2(-1)^N$$
we have
$$F_{2^{n+1}}=F_{2^n}L_{2^n}$$
$$L_{2^{n+1}}=L_{2^n}^2-2$$
and so
$$\begin{align}\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}&=\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^n}^2-4}\\\\&=\frac 32-\frac{F_{2^n}(L_{2^n}+2)-2F_{2^n}}{2(L_{2^n}-2)(L_{2^n}+2)}\\\\&=\frac 32-\frac{F_{2^n}L_{2^n}}{2(L_{2^{n}}^2-4)}\\\\&=\frac 32-\frac 12\cdot\frac{F_{2^{n+1}}}{L_{2^{n+1}}-2}\end{align}$$
