How to prove by Induction on k. using Fibonacci Numbers

[Spider]

Please guide me how to prove the last equation by induction.
Regards!

Conversely, any representation of the form (Zeckendorf's Theorem)
$$n=F_{k_{1}}+F_{k_{2}}+...+F_{k_{r}}, \quad k_{1}\gg k_{2}\gg ... \gg k_{r} \gg 0.$$
Implies that
$$ F_{k_{1}} \leq n < F_{k_{1}+1}, $$
because the largest possible value of $F_{k_{2}}+...+F_{k_{r}},$ when $k\gg k_{2}\gg...\gg k_{r}\gg0\:$is
$$F_{k-2}+F_{k-4}+...+F_{k\,mod\,2+2}=F_{k-1}-1, \quad \quad if\: k\geq2.$$
This last 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.

 

[Valkarie]

Proof by induction:Base case: When $k=2$, the equation becomes $F_{k-2}+F_{k-4}+...+F_{k\,mod\,2+2}=F_0-1=0$. Inductive step: Assume that the equation holds for some $k\geq2$. Then we have to show that it also holds for $k+1$.$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}$ (by inductive hypothesis)$= F_{k-1}+F_{k-3}+...+F_{k+1-2}$ (since $k+1 \geq 2$)$ = F_{k-1}+F_{k-2}+F_{k-4}+...+F_{k+1-2}$ (since $k+1-2 \geq 0$)$= F_{k+1-1}-1$ (by definition of Fibonacci numbers)Therefore, the equation holds for $k+1$. By mathematical induction, the equation holds for all $k \geq 2$.