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.

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.

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...