Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer Let $f_n$ be the Fibonacci numbers, defined by $f_0 = 0$ and $f_1 = 1$ and $f_n = f_{n-1} + f_{n-2}$ for all $n \ge 2$.

Show that $f_0 - f_1 + f_2 - \cdots - f_{2n-1} + f_{2n} = f_{2n-1} - 1$ when $n$ is a positive integer.

Just some homework help. Need to prove. Thank you in advance.
 A: I will give a less beautiful method, but it could be interesting and I think it's worth mentioning.
Let $F=\begin{bmatrix}1&1\\1&0\end{bmatrix}$, with the property that $F^n=\begin{bmatrix}f_{n+1}&f_{n}\\f_{n}&f_{n-1}\end{bmatrix}$ for $n\in\mathbb Z$.
Consider $(-F)^0+(-F)^1+(-F)^2+\cdots+(-F)^{2n}$. This is $(-F-I_2)^{-1}((-F)^{2n+1}-(-F)^0)$.
A few calculations give $(-F-I_2)^{-1}=\begin{bmatrix}-1&1\\1&-2\end{bmatrix}$. Comparing the entries in the first row and second column on both sides, we see that the sum $f_0-f_1+f_2-\cdots+f_{2n}$ equals
$$(-1)\cdot(-f_{2n+1})+1\cdot(-f_{2n}-1)=f_{2n-1}-1,$$
as desired.
Note that this method allows you to prove a closed form for such sums without knowing what the result should be.
A: Hint 1: Induction...
Hint 2: replace $f_n$ by their "closed" form, and see how you can calculate those sums.. It is easy...
Each hint leads to a different solution...
P.S. I assumed that $f_n$ is the  Fibonacci sequence, I am pretty sure it is...
A: In case you already now the formula for the sum of first $n$ Fibonacci numbers
$$\sum_{j=0}^n F_j = F_{n+2} - 1$$
you can use it as follows:
$F_0-F_1+F_2-F_3+\dots-F_{2n-1}+F_{2n}=$
$F_0+(F_2-F_1)+(F_4-F_3)+\dots+(F_{2n}-F_{2n-1})=$
$0+F_0+F_2+\dots+F_{2n-2}=$
$0+(F_0+F_1)+\dots+(F_{2n-4}+F_{2n-3})=$
$\sum_{k=0}^{2n-3} F_n = F_{2n-1}-1.$
A proof of the above formula for the sum of the first $n$ Fibonacci numbers can be found e.g. at proofwiki. 
