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 $$\begin{align}\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^{n+1}}-2}&=\frac 32-\frac 12\cdot\frac{F_{2^n}}{L_{2^n}-2}+\frac{F_{2^n}}{L_{2^n}^2-4}\\\\&=\frac 32-\frac{F_{2^n}(L_{2^n}+2)-2F_{2^n}}{2(L_{2^n}-2)(L_{2^n}+2)}\\\\&=\frac 32-\frac{F_{2^n}L_{2^n}}{2(L_{2^{n}}^2-4)}\\\\&=\frac 32-\frac 12\cdot\frac{F_{2^{n+1}}}{L_{2^{n+1}}-2}\end{align}$$