Using induction prove that: $$\sum_{i=0}^{2n} (-1)^iF_i = (F_{2n-1})-1$$
I am on the inductive step but am stuck with how to proceed.
This is what I have after the base case so far:
$$\sum_{i=0}^{2n+2} (-1)^iF_i = (F_{2n+2})-1$$ = $$(F_{2n+1})+F_{2n}-1$$ = $$(F_{2n-1})+F_{2n}+F_{2n}-1$$ I'm not sure how to simplify this out
Sketch: For $n=1$, the LHS becomes $$ F_0-F_1+F_2=0-1+1=0 $$ and the RHS becomes $$ F_1-1=1-1=0. $$ Since the LHS and RHS are both zero, they are equal.
Now, assume that the claim is true for $n=k$. Therefore $$ \sum_{i=0}^{2k}(-1)^iF_i=F_{2k-1}-1. $$ Consider the case where $n=k+1$. In this case, the LHS is $$ \sum_{i=0}^{2(k+1)}(-1)^iF_i=F_{2k+2}-F_{2k+1}+\sum_{i=0}^{2k}(-1)^iF_i. $$ Now, use the inductive hypothesis and break up your Fibonacci numbers. Recall that your goal is to get that the RHS is $F_{2(k+1)-1}-1=F_{2k+1}-1$, so break things up to try to get towards that expression.