Proof about specific sum of Fibonacci numbers Let $F_k$ denote the $k$-th Fibonacci number. Find a formula for and prove by induction that your formula is correct for all $n > 0$.
$$
(-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_n=\ ?
$$
I have tried finding several formulas but all of them were wrong or where working only when n>1 please help me if you can
 A: Let $A_n=\sum_{k=0}^{n}(-1)^{n-k} F_k$. Since:
$$ \frac{x}{1-x-x^2}=\sum_{k\geq 0} F_{k}\,x^k,\tag{1} $$
we have:
$$ A_n = [x^n]\left(\frac{1}{1+x}\cdot\frac{x}{1-x-x^2}\right),\tag{2} $$
but by partial fraction decomposition and $(1)$:
$$ \forall n\geq 1,\qquad A_n = [x^n]\left(1-\frac{1}{1+x}+\frac{x^2}{1-x-x^2}\right) = F_{n-1}-(-1)^n\tag{3}$$
hence:

$$ \forall{n\geq 1},\qquad \sum_{k=0}^{n}(-1)^k F_k = (-1)^n F_{n-1}-1.\tag{4} $$

A: Assuming, $ F_{0} = 0 \ and\ F_{1} = 1 $.
We have two properties of Fibonacci numbers link
$ F_{1} + F_{3} + .... + F_{2*n-1} = F_{2*n} $
$ F_{2} + F_{4} + .... + F_{2*n} = F_{2*n+1} -1 $
$(-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_{2*n}= F_{2*n+1} - F_{2*n} - 1$
Hope this will help you.
A: Using the famous closed form for the Fibonacci numbers and some power series, we can establish that your sum is the same as
$$\frac{(-2)^{-n} (1+\sqrt{5})^{-n-1} (2 (1+\sqrt{5}))^{2n}+(3+\sqrt{5}) (-4)^n-(5+\sqrt{5}) (-2 (1+\sqrt{5}))^n)}{\sqrt{5}}$$
Although this will probably not satisfy you, it is irrefutably a closed form for this sum.
It might be possible to transform it into a nicer one though...
A: Prove by induction that
$$\sum_{i = 0}^{n}(-1)^{i}F_i = (-1)^nF_{n-1}-1$$
assuming that you have $F_{-1}=-1, F_0=0, F_1=1$
A: Let 
$$
S_n=(-1)^0 F_0+(-1)^1 F_1+(-1)^2 F_2+\cdots+(-1)^n F_n.
$$
If $F_0=F_1=1$ then $S_n=(-1)^n F_{n-1}+1$.
