Sum of Sequence of Fibonacci Numbers

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

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

Proof
From the initial definition of Fibonacci numbers, we have:
 * $F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$

Proof by induction:

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

Basis for the Induction
$P(2)$ is the case:
 * $F_1 + F_2 = 2 = F_4 - 1$

which is seen to hold.

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:
 * $\displaystyle \sum_{j \mathop = 1}^k F_j = F_{k + 2} - 1$

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

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:
 * $\displaystyle \forall n \ge 2: \sum_{j \mathop = 1}^n F_j = F_{n + 2} - 1$