- \( F _ 0 = 0 \) and \( F _ 1 = 1 \)
- \( F _ n = F _ { n - 1 } + F _ { n - 2 } \) for every \( n \ge 2 \)

The Fibonacci Numbers

We define the \( n \)-th fibonacci number as follows

- \( F _ 0 = 0 \) and \( F _ 1 = 1 \)
- \( F _ n = F _ { n - 1 } + F _ { n - 2 } \) for every \( n \ge 2 \)

Sum Equals Jump 2 minus One

For every \( n \in \mathbb{ N } _ 0 \) we have
\[
F _ 0 + F _ 1 + \cdots + F _ n = F _ { n + 2 } - 1
\]

We prove it by induction, for the base case \( F \left( 0 + 2 \right) - 1 = \left( F _ 1 + F _ 0 \right) - 1 = 1 + 0 - 1 = 0 = F _ 0 \) as needed. Next assume it holds true for \( k \in \mathbb{ N } _ 0 \) and then we want to prove that \( F _ { k + 3 } - 1 = F _ 0 + \ldots + F _ { k + 1 } \), we can handle the sum with the inductive hypothesis, so that
\[
F _ 0 + \ldots + F _ k + F _ { k + 1 } = \left( F _ { k + 2 } - 1 \right) + F _ { k + 1 } = F _ { k + 3 } - 1
\]
as needed.