I'm stuck on proving this statement below due to that I can't seem to find a base case that is true:
Prove that $$\sum_{i=0}^{2n}(-1)^if(i) = f(0) - f(1) + f(2) - \cdots - f(2n-1) + f(2n) = f(2n-1)-1,$$ where $f(i)$ is the $i$th Fibonacci number.
I believe simple induction should work but I am unsure if I have to use Strong Induction.
For $n+1$ we have
$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=\sum_{i=0}^{2n}(-1)^if(i)-f(2n+1)+f(2n+2)=\\ =f(2n-1)-1-f(2n+1)+f(2n+2)$$
but
$$f(2n+2)=f(2n+1)+f(2n),\\ f(2n+1)=f(2n)+f(2n-1)$$ so,
$$f(2n-1)-f(2n+1)+f(2n+2)=f(2n+1)$$
and then
$$\sum_{i=0}^{2(n+1)}(-1)^if(i)=f(2n+1)-1=f[2(n+1)-1]-1$$