Prove by Induction on k. using Fibonacci Numbers $$F_{k-2}+F_{k-4}+...+F_{k\,mod\,2+2}=F_{k-1}-1, \quad \quad if\: k\geq2.$$
This equation is to prove by induction on $k;$ the left-hand side is zero when $k$ is $2$ or $3$. Therefore $k_{1}$ is the greedily chosen value described earlier, and the representation must be unique. 
Here is my Attempt. 
I have attempted to solve this problem using induction, Please if anyone confirms that my attempt is true for induction step. Or If someone helps me with this answer if anything goes wrong.
$$F_{k-2}+F_{k-4}+...+F_{k\,mod\,2+2}=F_{k-1}-1 \quad \quad if\,k\geq2$$
My Attempt:
Base case $k=2$
$$F_{2-2}+F_{2-4}+...+F_{2\,mod \,2+2}=F_{2-1}-1$$
As $2\, mod\,2=0$
therefore,
$$F_{0}+F_{-2}+...+F_{0+2}=F_{1}-1$$
$$F_{0}+F_{-2}+...+F_{2}=F_{1}-1$$
As $F_{0}=0,\,F_{-2}=-1,\,F_{2}=1,\,and\,F_{1}=1\,$therefore,
$$0+(-1)+...+1=1-1$$
$$0-1+...+1=1-1$$
$$0=0 \\ which\,\,is\,\,true\,\,the\,\,left\,\,hand\,\,side\,\,is\,\,zero\,\,when\,\,k\,\,is\,\,2.$$ 
Now the Induction Step: $k=k+1$ on left hand side
$$F_{k+1-2}+F_{k+1-4}+...+F_{k+1\,mod\,2+2}$$
$$F_{k-1}+F_{k-3}+...+F_{k+1\,mod\,2+2}$$
As we know that $F_{k-1}=F_{k+1}+F_{k}$ and $F_{k-3}=2F_{k+1}-3F_{k}$ therefore,
$$F_{k+1}+F_{k}+2F_{k+1}-3F_{k}+...+F_{k+1\,mod\,2+2}$$
$$(F_{k+1}+2F_{k+1})+(F_{k}-3F_{k})+...+F_{k+1\,mod\,2+2}$$
$$(3F_{k+1})+(-2F_{k})+...+F_{k+1\,mod\,2+2}$$
$$3F_{k+1}-2F_{k}+...+F_{k+1\,mod\,2+2}$$
 A: The question can be rewritten as
$$
F_{n-1}-1=\sum_{k=1}^{\left\lfloor\frac{n-2}2\right\rfloor}F_{2k+(n\bmod2)}\tag1
$$
For $n=2$ or $n=3$, the sum on the right side of $(1)$ is an empty sum and the left hand side of $(1)$ is $0$. For these two cases, $(1)$ holds.
Suppose that $(1)$ holds. Adding $F_n$ to both sides gives
$$
F_{n+1}-1=\sum_{k=1}^{\left\lfloor\frac{n}2\right\rfloor}F_{2k+(n\bmod2)}\tag2
$$
the left side follows by the recurrence $F_{n+1}=F_n+F_{n-1}$ and the right side follows because $F_{2\left\lfloor\frac{n}2\right\rfloor+(n\bmod2)}=F_n$ for both $n$ even and $n$ odd. Since $(2)$ is $(1)$ for $n\mapsto n+2$, $(1)$ follows for all $n\ge2$, even and odd.

We can combine the parallel inductions above as a single induction. Break $(1)$ into $P(m)$:
$$
\begin{align}
F_{2m-1}-1&=\sum_{k=1}^{m-1}F_{2k}\tag3\\
F_{2m}-1&=\sum_{k=1}^{m-1}F_{2k+1}\tag4
\end{align}
$$
$P(1)$ is simply
$$
\begin{align}
\overbrace{F_1-1\vphantom{\sum_{k=1}^0}}^0&=\overbrace{\sum_{k=1}^0F_{2k}}^0\tag5\\
F_2-1&=\sum_{k=1}^0F_{2k+1}\tag6
\end{align}
$$
Now suppose that $P(m)$ is true. Add $F_{2m}$ to both sides of $(3)$ and $F_{2m+1}$ to both sides of $(4)$. We get
$$
\begin{align}
F_{2m+1}-1&=\sum_{k=1}^{m}F_{2k}\tag7\\
F_{2m+2}-1&=\sum_{k=1}^{m}F_{2k+1}\tag8
\end{align}
$$
which is $P(m+1)$.
This completes the induction.
A: I have attempted to solve this problem using induction, Please if anyone confirms that my attempt is true for induction step. Or If someone helps me with this answer if anything goes wrong.
$$F_{k-2}+F_{k-4}+...+F_{k\,mod\,2+2}=F_{k-1}-1 \quad \quad if\,k\geq2$$
My Attempt:
Base case $k=2$
$$F_{2-2}+F_{2-4}+...+F_{2\,mod \,2+2}=F_{2-1}-1$$
As $2\, mod\,2=0$
therefore,
$$F_{0}+F_{-2}+...+F_{0+2}=F_{1}-1$$
$$F_{0}+F_{-2}+...+F_{2}=F_{1}-1$$
As $F_{0}=0,\,F_{-2}=-1,\,F_{2}=1,\,and\,F_{1}=1\,$therefore,
$$0+(-1)+...+1=1-1$$
$$0-1+...+1=1-1$$
$$0=0 \\ which\,\,is\,\,true\,\,the\,\,left\,\,hand\,\,side\,\,is\,\,zero\,\,when\,\,k\,\,is\,\,2.$$ 
Now the Induction Step: $k=k+1$ on left hand side
$$F_{k+1-2}+F_{k+1-4}+...+F_{k+1\,mod\,2+2}$$
$$F_{k-1}+F_{k-3}+...+F_{k+1\,mod\,2+2}$$
As we know that $F_{k-1}=F_{k+1}+F_{k}$ and $F_{k-3}=2F_{k+1}-3F_{k}$ therefore,
$$F_{k+1}+F_{k}+2F_{k+1}-3F_{k}+...+F_{k+1\,mod\,2+2}$$
$$(F_{k+1}+2F_{k+1})+(F_{k}-3F_{k})+...+F_{k+1\,mod\,2+2}$$
$$(3F_{k+1})+(-2F_{k})+...+F_{k+1\,mod\,2+2}$$
$$3F_{k+1}-2F_{k}+...+F_{k+1\,mod\,2+2}$$
