Negafibonacci Identity Consider the sequence of the negafibonacci numbers: $$..., \overbrace{-8}^{F_{-6}}, \overbrace5^{F_{-5}}, \overbrace{-3}^{F_{-4}}, \overbrace2^{F_{-3}}, \overbrace{-1}^{F_{-2}}, \overbrace1^{F_{-1}}$$
I have discovered that the following identity holds for integers I've tried:

$$\sum_{n=1}^kF_{-n}=1-F_{1-k}$$

How could this be shown for any $k\in\mathbb{N}$? 
 A: We want to prove that $$\sum_{k=1}^n F_{-k} = 1-F_{1-n}$$


*

*We know that $F_0 = 0$, $F_1 = 1$, and that therefore $F_{-1} = 1$.

*Hence the base case for $n=1$ holds: $$F_{-1} = 1-F_{0} =  1$$

*In the inductive step, suppose the statement holds for $n$. We'll show it holds for $n+1$. Indeed,
$$\begin{align*}\sum_{k=1}^{n+1} F_{-k} &= \sum_{k=1}^n F_{-k}\; + F_{-(n+1)} & \text{\{extract last term\}}\\  &= \sum_{k=1}^n F_{-k}\; + F_{-(n-1)} - F_{-n}&\text{\{fibonacci relation: } F_{-(n+1)}+F_{-n}=F_{-(n-1)}{\}}\\
&= \left[1-F_{1-n}\right] + F_{-(n-1)} - F_{-n}&\text{\{inductive hypothesis\}}\\
&= 1 - F_{-n}\\
&= 1 - F_{1-(n+1)}
\end{align*} $$
which was to be shown.
A: Here is an answer based upon generating functions.  But at first note that the identity
\begin{align*}
\sum_{k=1}^nF_{-k}=1-F_{1-n}\qquad\qquad n\geq 1\tag{1}
\end{align*}
gives for $n=1$: $F_{-1}=1-F_{0}$ and so we have to add the initial condition
\begin{align*}
F_{0}=0
\end{align*}
in order to fully specify the identity.

We use the generating function of the Fibonacci numbers:
  \begin{align*}
\frac{x}{1-x-x^2}=x+x^2+2x^3+5x^4+8x^5+13x^6+\cdots
\end{align*}
  and the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
  \begin{align*}
F_n=[x^n]\frac{x}{1-x-x^2}\qquad\qquad n\geq 1
\end{align*}

Since $F_{-n}=(-1)^{n+1}F_n$ with $F_n$ the Fibonacci numbers, the claim (1) can be written as
\begin{align*}
\sum_{k=1}^n(-1)^{k+1}F_k=1-(-1)^nF_{n-1}\qquad\qquad n\geq 1\tag{2}
\end{align*}

We start with the left-hand side of (2) and obtain
  \begin{align*}
\color{blue}{\sum_{k=1}^n(-1)^{k+1}F_k}&=\sum_{k=1}^n (-1)^{k+1}[x^k]\frac{x}{1-x-x^2}\tag{3}\\
&=\sum_{k=1}^n[x^k]\frac{x}{1+x-x^2}\tag{4}\\
&=[x^0]\frac{x}{1+x-x^2}\sum_{k=1}^n\frac{1}{x^k}\tag{5}\\
&=[x^0]\frac{x}{1+x-x^2}\cdot\frac{1-x^{n}}{x^n(1-x)}\tag{6}\\
&=[x^n]\frac{x}{(1+x-x^2)(1-x)}\tag{7}\\
&=[x^n]\left(\frac{x}{1-x}-\frac{x^2}{1+x-x^2}\right)\tag{8}\\
&=1-[x^n]\frac{x^2}{1+x-x^2}\\
&=1-(-1)^n[x^n]\frac{x^2}{1-x-x^2}\\
&\color{blue}{=1-(-1)^nF_{n-1}}
\end{align*}
  and the claim follows.

Comment:


*

*In (3) we write $F_k$ using the  coefficient of operator.

*In (4) we note that multiplication of the coefficient of $x^n$ with $(-1)^n$ means replacing $x$ with $-x$.

*In (5) we use the linearity of the coefficient of operator and apply the rule  $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$.

*In (6) we use the *finite geometric series formula and do some simplifications.

*In (7) we again apply the rule $[x^{p+q}]A(x)=[x^p]x^{-q}A(x)$ as we did in (5) and observe that we can skip $x^{n+1}$ in the numerator, since it does not contribute to $[x^n]$.

*In (8) we do a partial fraction decomposition as preparation for the final steps.
A: Since
$F_{-n}=(-1)^{n+1}F_n$,
your identity
is equivalent to
$\sum_{n=1}^k(-1)^{n+1}F_{n}=1-(-1)^{k}F_{k-1}
$.
These two identities
can be used:
$$\sum_{i=0}^{n-1}F_{2i+1} = F_{2n}
\\\sum_{i=1}^{n}F_{2i} = F_{2n+1}-1
.$$
Rewrite the first as
$\sum_{i=1}^{n}F_{2i-1} = F_{2n}
$.
Subtracting this
from the second,
$$\sum_{i=1}^{n}(F_{2i}-F_{2i-1})
 = F_{2n+1}-F_{2n}-1
\implies\sum_{i=1}^{2n} (-1)^{i}F_i
=F_{2n-1}-1
$$
or
$$\sum_{i=1}^{2n} (-1)^{i+1}F_i
=1-F_{2n-1}
=1-(-1)^{2n}F_{2n-1}
$$
or
$$\sum_{i=1}^{2n} F_{-i}
=1-F_{2n-1}
=1-F_{1-2n}
=1-(-1)^{2n}F_{1-2n}.
$$
So your identity is true
for even $k$.
To show this for odd $k$,
split the last term from
$\sum_{i=1}^{2n} (-1)^{i}F_i
=F_{2n-1}-1
$.
This becomes
$$\sum_{i=1}^{2n-1} (-1)^{i}F_i+(-1)^{2n}F_{2n}
=F_{2n-1}-1
$$
or
$$\sum_{i=1}^{2n-1} (-1)^{i}F_i
=F_{2n-1}-1-F_{2n}
=-F_{2n-2}-1
=(-1)^{2n-1}F_{2n-2}-1
$$
and this will give your identity
for odd $k$.
