Recurrence Formula for Sum of Sequence of Fibonacci Numbers

Theorem
Let $g_n$ be the sum of the first $n$ Fibonacci numbers.

Then:
 * $\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$

Proof
Let $F_n$ be the $n$th Fibonacci number.

By definition:
 * $F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3, \ldots$

Hence, $g_n$, the sum of the first $n$ Fibonacci numbers is
 * $g_0 = 0, g_1 = 1, g_2 = 2, g_3 = 4, g_4 = 7, \ldots$

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{j \mathop = 0}^n F_j = g_n = g_{n - 1} + g_{n - 2} + 1$

Basis for the Induction
$P(2)$ is the case $g_2 = g_1 + g_0 + 1$, which holds:
 * $2 = 1 + 0 + 1$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 0}^k F_j = \sum_{j \mathop = 0}^{k - 1} F_j + \sum_{j \mathop = 0}^{k - 2} F_j + 1$

Then we need to show:
 * $\displaystyle \sum_{j \mathop = 0}^{k + 1} F_j = \sum_{j \mathop = 0}^k F_j + \sum_{j \mathop = 0}^{k - 1} F_j + 1$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \ge 2: g_n = g_{n - 1} + g_{n - 2} + 1$

where $g_n$ is the sum of the first $n$ Fibonacci numbers.