How can we show that the following (recursive) statement, where $\displaystyle F_n$ is the $\displaystyle n$-th Fibonacci number, and $\displaystyle n \geq 1$, is true?

$\displaystyle \sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$

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As this should be shown to be valid for every $\displaystyle n \in \mathbb{N}$, I believe we can use induction. The least $\displaystyle n$ which has to satisfy the above equation is $\displaystyle 1$, so we take the case when $\displaystyle n=1$ as basis. But,

$\displaystyle \sum_{i=0}^{\lfloor \frac{1-1}{2} \rfloor} F_{1-2i}

\stackrel{\text{?}}{=}

F_{1+1}-1$

$\displaystyle \sum_{i=0}^{0} F_{1-2i}

\stackrel{\text{?}}{=}

F_{2}-1$

$\displaystyle F_{1}

\neq

F_{2}-1$, because $\displaystyle F_1=1$, and $\displaystyle F_2-1=1-1=0$.

However, if we take the case when $\displaystyle n=2$ as basis, then the above statementistrue:

$\displaystyle \sum_{i=0}^{\lfloor \frac{1}{2} \rfloor} F_{2-2i}=\sum_{i=0}^{0} F_{2-2i}=F_2=1=$

$\displaystyle F_{2+1}-1=F_3-1=2-1=1$

Is this problem wrongly posed? Should we show that $\displaystyle \sum_{i=0}^{\lfloor \frac{n-1}{2} \rfloor} F_{n-2i}=F_{n+1}-1$ is true for every $\displaystyle n \geq 2$ instead of $\displaystyle n \geq 1$?

And my next question would be how to proceed with the inductive step? I know we should use the fact that $\displaystyle F_n=F_{n-1}+F_{n-2}$ somewhere, but so far I couldn't show that the above is true for $\displaystyle k+1$ if we suppose it is valid form some $\displaystyle k \in \mathbb{N}$.

I'll be immensely grateful for any help.