# 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:

 $\ds F_0$ $=$ $\ds 0$ $\ds$ $=$ $\ds 1 - 1$ $\ds$ $=$ $\ds F_2 - 1$

which is seen to hold.

### Basis for the Induction

$\map P 1$ is the case:

 $\ds F_1$ $=$ $\ds 1$ $\ds$ $=$ $\ds 2 - 1$ $\ds$ $=$ $\ds F_3 - 1$

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:

 $\ds \sum_{j \mathop = 1}^{k + 1} F_j$ $=$ $\ds \sum_{j \mathop = 1}^k F_j + F_{k + 1}$ $\ds$ $=$ $\ds F_{k + 2} - 1 + F_{k + 1}$ Induction Hypothesis $\ds$ $=$ $\ds F_{k + 3} - 1$ Definition of Fibonacci Number

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

Therefore:

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

$\blacksquare$

## 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 $F_0 = 0$.