Sum of Sequence of Fibonacci Numbers

Theorem
Let $F_n$ denote the $n$th Fibonacci number.

Then:
 * $\ds \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$

Proof
Proof by induction:

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\ds \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$

$\map P 0$ is the case:

which is seen to hold.

Basis for the Induction
$\map P 1$ is the case:

which is seen to hold.

This is our basis for the induction.

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

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

Then we need to show:
 * $\ds \sum_{j \mathop = 1}^{k + 1} F_j = F_{k + 3} - 1$

Induction Step
This is our induction step:

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

Therefore:
 * $\ds \forall n \in \Z_{\ge 0}: \sum_{j \mathop = 0}^n F_j = F_{n + 2} - 1$

Also presented as
This can also be seen presented as:
 * $\ds \sum_{j \mathop = 1}^n F_j = F_{n + 2} - 1$

which is seen to be equivalent to the result given, as $F_0 = 0$.