Fibonacci sequence proof. Given the Fibonacci Sequence $F_{0}=0,F_{1}=1,F_{n+2}=F_{n+1}+F_{n+1};\;n\geq0$
Prove the following:
$\sum_{i=1}^{n}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{n+2}}{2^{n}}$
Base case:
$\frac{F_{0}}{2^{0}}= 1-\frac{F_{2}}{2^{0}}\\ \frac{0}{1} = 1-\frac{1}{1} \\ 0 =0$
Which is true.
Now, for the inductive step with $n=k$ we have
$\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+2}}{2^{k}}$
We want to prove then that for $n=k+1$
$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$
We rewrite the LHS such that
$\frac{F_{k}}{2^{k+1}}+\sum_{i=1}^{k}\frac{F_{i-1}}{2^{i}} = 1-\frac{F_{k+3}}{2^{k+1}}$
From the inductive hypothesis it follows that
$\frac{F_{k}}{2^{k+1}}+ 1-\frac{F_{k+2}}{2^{k}}= 1-\frac{F_{k+3}}{2^{k+1}}$
I get stuck here. Any hint will be appreciated.

Since $\sum_{i=1}^k\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}$,$$\sum_{i=1}^{k+1}\frac{F_{i-1}}{2^i}=1-\frac{F_{k+2}}{2^k}+\frac{F_k}{2^{k+1}}=1-\frac{2F_{k+2}-F_k}{2^{k+1}},$$so you just need to prove $2F_{k+2}-F_k=F_{k+3}$. With $a:=F_k,\,b:=F_{k+1}$ we get$$F_{k+2}=a+b,\,F_{k+3}=a+2b\implies2F_{k+2}-F_k=2(a+b)-a=a+2b=F_{k+3}.$$
