Prove Fibonacci identity by induction

I am having trouble with the induction step of this proof, any nudges in the right direction or pointers where my reasoning is wrong are greatly appreciated.

I should prove the following equality:

For all $n, n_1 \in \mathbb N, n_1 \ge 4, F_0 = 1, F_1 = 2, F_n = F_{n-1} + F_{n-2}$ $$F_{n_1-1} = 2 + \sum_{i=0}^{n_1-3} F_i$$

My work so far:

Base case $n_1 = 4$: $$F_3 = 2 + \sum_{i=0}^{1} F_i = 2 + F_0 + F_1$$ $$= 2 + 1 + 2$$ $$=5$$

IA: Assume for particular $k \in \mathbb N, k \ge 4$ $$F_{k-1} = 2 + \sum_{i+0}^{k-3} F_i$$

IS: Here I should show that this applies for $k+1$. $$F_{(k+1)-1} = 2 + \sum_{i=0}^{(k+1)-3} F_i$$ $$=2 + \sum_{i=0}^{k-2} F_i$$ And this is where I'm a bit stuck. I think I need to show that this is the same thing as the sum for k, plus a bit...right? And that "bit" should basically be enough to get us to the next term in the Fibonacci sequence?

Because we defined the nth Fibonacci term as $F_n = F_{n-1} + F_{n-2}$, I broke up the sum for $k+1$ like so: $$(2 + \sum_{i = 0}^{((k+1)-1) - 3}F_i) + (2 + \sum_{i=0}^{((k+1)-2)-3}F_i)$$ $$= (2 + \sum_{i = 0}^{k-3}F_i) + (2 + \sum_{i=0}^{k-4}F_i)$$ And then I can see that the first operand is actually the sum for $k$. But here I feel stuck. How do I show that the second operand is enough to get us to the term after $F_{k-1}$?

Answer

Let $F_0 = 0, F_1 = 1$ and $F_{n+1} = F_n + F_{n-1}$. Then $$F_{n+2}-1 = \sum_{k=1}^n F_k.$$ I'm guessing you want to prove something along these lines. That's not too bad via induction.

Base Case: Exercise.

Induction Step: Suppose $F_{n+2} -1 = \sum_{k=1}^n F_k$ for a natural $n$. Observe $$\sum_{k=1}^{n+1} F_k = \sum_{k=1}^n F_k + F_{n+1} = (F_{n+2}-1)+F_{n+1} = F_{n+3} - 1.$$ That's it.